Category Archives: IITJEE Mains

Derivatives part 14: IITJEE maths tutorial problems for practice

This is part 14 of the series

Question 1:

Let f(x)= \sqrt{x-1} + \sqrt{x+24-10\sqrt{x-1}} for x<26 be a real valued function. Then, find f^{'}(x) for 1<x<26:

Answer 1:

Consider (\sqrt{x-1}-5)^{2} = x-1+25-10\sqrt{x-1} = x+24 -10\sqrt{x-1} so that we have


Hence, f(x) = \sqrt{x-1} + \sqrt{x-1}-5 = 2\sqrt{x-1}-5 when 1<x<26

Hence, f^{'}{x} = \frac{-2}{2\sqrt{x-1}} = -\frac{1}{\sqrt{x-1}}

Question 2:

Let 3f(x) - 2 f(\frac{1}{x})=x, then find f^{'}(2).

Answer 2:

Given that 3f(x) - 2 f(\frac{1}{x})=x….call this I.

Also, from above, we get 3f(\frac{1}{x}) - 2 f(x)= \frac{1}{x}…call this II.

so we get 6f(x)-4f(\frac{1}{x})=2x….call this I’

and 9f(\frac{1}{x})-6f(x) = 9/x…call this II’.

5f(\frac{1}{x})=2x+ \frac{9}{x} and hence, f^{'}(1/x) = \frac{2x}{5} + \frac{9}{5x}

Also, again 3f(x)-2f(1/x)=x….A


So, we now we get the following two equations:



so, now we have 5f(x) = 3x + \frac{2}{x} so that we get f(x) = \frac{3}{5}x+\frac{2}{5x} andf(1/x) = \frac{2}{5}x+\frac{9}{5x}

so f^{'}(x) = \frac{3}{5} + \frac{2}{5}\frac{-1}{x^{2}} = \frac{3}{5} - \frac{1}{10}=\frac{1}{2}

Question 3:

If x= \frac{a(1-t^{2})}{1+t^{2}} and y = \frac{2bt}{1+t^{2}}, then find \frac{dy}{dx}.

Answer 3:

Given that x = \frac{a(1-t^{2})}{1+t^{2}} where a is a parameter (constant) and t is a variable.

Let t=\tan{\theta} so that x = \frac{a(1-\tan^{2}{\theta})}{1+\tan^{2}{\theta}} = a \cos{2\theta}

so that y = \frac{2bt}{1+t^{2}}=b \sin{2\theta}

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{2b\cos{2\theta}}{-2a\sin{2\theta}}=- \frac{b}{a\tan{2\theta}}

so that we have

\frac{dy}{dx} = - \frac{b(1-t^{2})}{a}

Question 4:

If y = \arccos{\frac{x-x^{-1}}{x+x^{-1}}} then find \frac{dy}{dx}

Answer 4:

Given that \arccos{\frac{x^{2}-1}{x^{2}+1}} = \arccos{\frac{1-x^{2}}{1+x^{2}}} and put x=\tan{\theta}

\frac{1-\tan^{2}{\theta}}{1+\tan^{2}{\theta}}=\cos{2\theta} so that y = \arccos {\cos{2\theta}}=2\theta

\frac{dy}{dx} = 2\frac{d}{dx}(\arctan{x})=\frac{2}{1+x^{2}} which is the required answer.

Question 5:

If \arcsin{(\frac{x^{2}-y^{2}}{x^{2}+y^{2}})}=a, where a is a parameter, then find \frac{dy}{dx}.

Answer 5:

Given that a=\arcsin{(\frac{x^{2}-y^{2}}{x^{2}+y^{2}})} so that \sin{a} = \frac{x^{2}-y^{2}}{x^{2}+y^{2}}

(x^{2}+y^{2})\sin{a} = x^{2}-y^{2}

Differentiating both sides w.r.t. x, we get

2x\sin{a} + \sin{a} ,2y.\frac{dy}{dx}=2x-2y\frac{dy}{dx}


\frac{dy}{dx} = \frac{2x(1-\sin{a})}{1+\sin{a}}=\frac{x}{y}\times \frac{2y^{2}}{2x^{2}}=\frac{y}{x}

Question 6:

If y = cot^{-1}{(\sqrt{\frac{1+x}{1-x}})} then find \frac{dy}{d(\arccos{x})}.

Answer 6:

Given that y = cot^{-1}(\sqrt{(\frac{1+x}{1-x})}) so that y = \arctan{(\sqrt{(\frac{1-x}{1+x})})} = \arctan{(cot {(2\theta)})} where x=\tan^{2}{\theta} so that \frac{d\theta}{dx} = \frac{1}{1+x^{2}}

and \sec^{2}{y}.\frac{dy}{dx} = - cosec^{2}{(2\theta)}.2\frac{d\theta}{dx}

Let f=\arccos{x} so that \frac{df}{dx} = - \frac{1}{\sqrt{1-x^{2}}}

Now, note that \sec^{2}{y} = cosec^{2}{2\theta} so we get the following simplification:

\frac{dy}{dx} = - \frac{2}{1+x^{2}}

Now, \frac{dy}{df} = -\frac{\frac{dy}{dx}}{\frac{df}{dx}}= \frac{2\sqrt{1-x^{2}}}{1+x^{2}}


Nalin Pithwa

Derivatives: part 13: IITJEE Math tutorial problems for practice

Question 1:

Let f(x) be a differentiable function w.r.t. x at x=1 and \lim_{h \rightarrow 0} \frac{1}{h}f(1+h)=5, then evaluate f^{'}(1)

Solution 1:

By definition, derivative of a function f(x) is f^{'}(x) = \lim_{t \rightarrow x}\frac{f(t)-f(x)}{t-x}, where let us substitute t-x=h, t=x+h, as t \rightarrow x, then h \rightarrow 0

So that above expression is equal to \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}

f^{'}(1) = \lim_{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} exists and can be evaluated if we know the value of the function f(x) at x=1.

Question 2:

If x\sqrt{1+y} + y \sqrt{1+x}=0, then find \frac{dy}{dx}.

Answer 2:

Given that x\sqrt{1+y} + y\sqrt{1+x}=0

Taking derivative of both sides w.r.t. x, we get the following equation:

\sqrt{1+y} \times 1 + \frac{x}{2\sqrt{1+y}}.\frac{dy}{dx} + \sqrt{1+x}\frac{dy}{dx}+\frac{y}{2\sqrt{1+x}}.1=0

\sqrt{1+y}+ \frac{y}{2\sqrt{1+x}}+\frac{dy}{dx} \times (\frac{x}{2\sqrt{1+y}}+\sqrt{1+x})=0

This further simplifies to :

\frac{dy}{dx}. (\frac{x+2\sqrt{(1+x)(1+y)}}{2\sqrt{1+y}}) = \frac{2\sqrt{(1+x)(1+y)}+y}{2\sqrt{1+x}}

\frac{}{} = 2 \times \sqrt{\frac{1+y}{1+x}} \times \frac{2\sqrt{(1+x)(1+y)+y}}{x+2\sqrt{(1+x)(1+y)}}

But, we already know that x\sqrt{1+y}=-y\sqrt{1+x} so that \sqrt{\frac{1+y}{1+x}} = - \frac{y}{x}

\frac{dy}{dx} = - \frac{2y}{x} \times \frac{2\sqrt{(1+x)(1+y)}+y}{x+2\sqrt{(1+x)(1+y)}}

\frac{dy}{dx}= -2 \times \frac{2y\sqrt{(1+x)(1+y)}+y^{2}}{x^{2}+2x\sqrt{(1+x)(1+y)}}

\frac{dy}{dx}=-2. \frac{2x(1+y)+y^{2}}{x^{2}-2y(1+y)} is the desired answer.

You can see how ugly it looks. Is there any way to simplify above? Let us give it one more shot. As follows:

Given that x\sqrt{1+y} + y\sqrt{1+x}=0 Hence, x^{2}(1+y)=y^{2}(1+x) so that


(x+y)(x-y) = y^{2}x-x^{2}y=xy(y-x). If x \neq y, then

x+y= -xy. Taking derivative of both sides w.r.t. x, we get:

1+\frac{dy}{dx} = y(-1)+(-x)\frac{dy}{dx}

(1+x)\frac{dy}{dx} = -1-y

\frac{dy}{dx} = - \frac{1+y}{1+x}= - \frac{y^{2}}{x^{2}} which is such an elegant answer 🙂

Question 3:

If x^{y}.y^{x}=c, where c is a parameter constant, then find \frac{dy}{dx} at (e,e).

Solution 3:

Let u=x^{y} and v=y^{x}.

Taking logarithm of both sides:

\log {u} = y \log {x} and \log {v} = x \log{y}.

Consider the LHS equation:

Taking derivative of both sides w.r.t.x, we get:

\frac{1}{u}\frac{du}{dx} = \frac{y}{x} + \frac{dy}{dx}. \log{x}

\frac{1}{u}. \frac{du}{dx} = \frac{y}{x} + \frac{dy}{dx} (\log{x})

\frac{du}{dx} = x^{y} \times (\frac{y}{x}+(\log{y}).\frac{dy}{dx})

\log {v} = x \log{y}

\frac{1}{v}\frac{dv}{dx} = \frac{x}{y}\frac{dy}{dx} + (\log{y})

\frac{dv}{dx} = y^{x} \times (\frac{x}{y}\frac{dy}{dx}+ \log{y}).

Also, x^{y}\frac{dv}{dx} + y^{x}\frac{du}{dx}=0

x^{y}y^{x} \times (\frac{x}{y}\frac{dy}{dx}+\log{y}) + y^{x}.x^{y}. (\frac{y}{x}+(\log{x}).\frac{dy}{dx}) =0

x^{y}y^{x} \times (\frac{dy}{dx}(\frac{x}{y}+\log{x})+\log{y}+\frac{y}{x})=0 Now substitute (x,y)=(e,e) and get the required answer.

x^{y}y^{x} \times (\frac{dy}{dx}(\frac{x}{y})+\log{x}) = - x^{y}y^{x}(\frac{y}{x}+\log{y})

\frac{dy}{dx} (\frac{x}{y}+\log{x})=-(\frac{y}{x}+\log{y})

Substituting (x,y) = (e,e)

Hence, then, (\frac{dy}{dx})_(e,e) = - \frac{e/e + log e}{e/e + \log {e}}=-1 is the desired answer.

Question 4:

Find \frac{d}{dx}(\tan(\arctan{x} + cot^{-1}(x+1))

Answer 4:

Consider \tan(A+B) = \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}

Subsituting A= \arctan{x} and B=cot^{-1}(x+1), we get the following:

\tan{(A+B)} = \frac{tan(\arctan{x}+tan(cot^{-1}(x+1)))}{1-x. tan(cot^{-1}(x+1))} = \frac{x+\frac{1}{(x+1)}}{1-\frac{x}{x+1}}

which in turn equals \frac{x+\frac{1}{x+1}}{1-\frac{x}{x+1}}=\frac{x^{2}+x+1}{1} noting that \arctan{(\frac{1}{x+1})}=cot^{-1}(x+1)

Hence, the answer is \frac{d}{dx}(x^{2}+x+1)=2x+1

Question 5:

If y = \arctan{\sqrt{\frac{1+\sin{x}}{1-\sin{x}}}}, find \frac{dy}{dx}

Solution 5:

Given that \tan{y} = \sqrt{\frac{1+\sin{x}}{1-\sin{x}}}

\tan^{2}{y} = \frac{1+\sin{x}}{1-\sin{x}}. Taking derivative of both sides w.r.t. x,

2(\tan{y})/ \sec^{2}{y}\frac{dy}{dx} = \frac{(1-\sin{x})(\cos{x})}{(1-\sin{x})^{2}}

2\tan{y}\sec^{2}{y}\frac{dy}{dx} = \frac{\cos{x}-\sin{x}\cos{x}+\cos{x}+\sin{x}\cos{x}}{(1-\sin{x})^{2}} which in turn equals


But, \tan^{2}{y} = \frac{1+\sin{x}}{1-\sin{x}}

so that \sec^{2}{y}=1+\tan^{2}{y}=1+\frac{1+\sin{x}}{1-\sin{x}} = \frac{2}{1-\sin{x}}

Hence, 2. \sqrt{\frac{1+\sin{x}}{1-\sin{x}}} \times \frac{2}{1-\sin{x}} \times \frac{dy}{dx} = \frac{2\cos{x}}{(1-\sin{x})^{2}}

Hence, \frac{dy}{dx} = \frac{1}{2} \times \frac{\cos{x}}{\sqrt{(1+\sin{x})(1-\sin{x})}} = \frac{\cos{x}}{2\sqrt{1-\sin^{2}{x}}} = \frac{1}{2}

Question 6:

Find \frac{d}{dx}cot^{-1}(\frac{1+\sqrt{1+x^{2}}}{x})

Solution 6:

Let y = cot^{-1} (\frac{1+\sqrt{1+x^{2}}}{x})

Put x = \sin{\theta} so that \sqrt{1-x^{2}}=\sqrt{1-\sin^{2}{\theta}}=\cos{\theta}

\frac{1+\cos{\theta}}{\sin{\theta}} = \frac{2\cos^{2}(\theta/2)}{2\sin{\theta/2}\cos{\theta/2}} = cot (\theta/2)

cot^{-1}(cot{(\theta/2)}) = \theta/2


\frac{dy}{dx} = \frac{1}{2}\frac{d}{dx}( \arcsin {x} )=\frac{1}{2\sqrt{1-x^{2}}} where |x|<1

Question 7:

If y = \arcsin{(\frac{2x}{1+x^{2}})}+sec^{-1}(\frac{1+x^{2}}{1-x^{2}}). Find \frac{dy}{dx}.

Solution 7:

Let x = \tan{\theta} so that \frac{2x}{1+x^{2}} = \frac{2\tan{\theta}}{1+\tan^{2}{\theta}} = \sin{2\theta}

so that \arcsin{(\frac{2x}{1+x^{2}})} = \arcsin{\sin{2\theta}}=2\theta

We now have \frac{1-x^{2}}{1+x^{2}} = \frac{1-\tan^{2}{\theta}}{1+\tan^{2}{\theta}} = \cos{2\theta}

so that sec^{-1}{(\frac{1+x^{2}}{1-x^{2}})} = sec^{-1}(sec {(2\theta)}) =2 \theta

so the desired answer is 4\frac{d\theta}{dx}=\frac{4}{1+x^{2}}

Question 8:

If y= \arcsin{(\frac{\sqrt{1+x}+\sqrt{1-x}}{2})} then find \frac{dy}{dx}

Solution 8:

Given that \sin{y} = \frac{\sqrt{1+x}+\sqrt{1-x}}{2}

2\cos{y}\frac{dy}{dx} = \frac{1}{2\sqrt{1+x}} \times 1 + \frac{1}{2\sqrt{1-x}} \times (-1)

2\cos{y} \frac{dy}{dx} = \frac{\sqrt{1-x}-\sqrt{1+x}}{2\sqrt{1-x^{2}}}

2\cos{y} \frac{dy}{dx} = \frac{1-x-(1+x)}{(2\sqrt{1-x^{2}})(\sqrt{1-x}+\sqrt{1+x})} = - \frac{2x}{(2\sqrt{1-x^{2}})(\sqrt{1-x}+\sqrt{1+x})}

2\cos{y} \frac{dy}{dx} = - \frac{x}{(\sqrt{1-x^{2}})(2\sin{y})}

\frac{dy}{dx} = - \frac{x}{2\sqrt{1-x^{2}} (\sin{2y})} but \sin{2y} = \frac{\sqrt{1+x}+\sqrt{1-x}}{2} and \cos{y} = \frac{\sqrt{1+x}-\sqrt{1-x}}{2}

so now we have \sin{2y} = \frac{2}{4} (1+x-(1-x)) = x

Hence, we get \frac{dy}{dx} = - \frac{x}{2\sqrt{1-x^{2}}(x)}=-\frac{1}{2\sqrt{1-x^{2}}}.

Question 9:

If g(x) = x^{2}+2x+3f(x) and f(0)=5 and \lim_{x \rightarrow 0} \frac{f(x)-5}{x}=4, then evaluate g^{'}(0).

Solution 9:

We have g^{'}{(x)} = 2x+2+3f^{'}(x) and hence, g^{'}{(0)}=2+3f^{'}{(0)}

By definition of derivative, we have f^{'}{(x)}= \lim_{t \rightarrow x} \frac{f(t)-f(x)}{t-x} where let us say t-x=h so that t \rightarrow x, and h  \rightarrow 0

f^{'}(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}

\lim_{h \rightarrow 0} \frac{f(h)-5}{h}=4 and f(0)=5 and hence, f^{'}(0)=4.

Hence, g^{'}(0)=2+3 \times 4=14

Question 10:

If \tan{y} = \frac{2t}{1-t^{2}}, and \sin{x}=\frac{2t}{1+t^{2}}, then find \frac{d^{2}y}{dx^{2}}.

Answer 10:

Let t = \tan{\theta} and hence, \frac{2t}{1-t^{2}} = \frac{2\tan{\theta}}{1-\tan^{2}{\theta}} = \tan{2\theta}

Hence, \tan{y} = \tan{2\theta}

so that \sec^{2}{y} \frac{dy}{dx} = \sec^{2}{(2\theta)}.2.\frac{d\theta}{dx}

Let t=\tan{\theta} so that \theta = \arctan{t} and \frac{d\theta}{dx}=\frac{1}{1+t^{2}} \times \frac{dt}{dx}

Hence, we get (\sec^{2}{y})\frac{dy}{dx} = \sec^{2}(2\theta) \times \frac{2}{(1+t^{2})}.\frac{dt}{dx} so that

\sec^{2}{(2\theta)}\frac{dy}{dx} = \sec^{2}{(2\theta)} \times \frac{2}{1+t^{2}} \times \frac{dt}{dx}

Hence, \frac{dy}{dx} = \frac{2}{(1+t^{2})}\frac{dt}{dx}

Hence, \sin{x}=\frac{2t}{1+t^{2}} = \frac{2\tan{\theta}}{1+\tan^{2}{(\theta)}}=2 \frac{\sin{\theta}}{\cos{\theta}} \frac{\cos^{2}{\theta}}{1}

Hence, \sin{x} = \sin{2\theta} and hence x=2\theta and so t=\tan{\theta} hence, \theta=\arctan{t}

x =\arctan{t} so that t=\tan{x}

\frac{dt}{dx} = \frac{1}{1+x^{2}}….call this A.

\frac{dy}{dx} = \frac{2}{1+t^{2}} \frac{dt}{dx}…call this B.

\frac{dy}{dx} = \frac{2}{(1+t^{2})(1+x^{2})}

\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(\frac{2}{(1+t^{2})(1+x^{2})}) = \frac{2}{(1+t^{2})} \frac{d}{dx}(\frac{1}{1+x^{2}}) + \frac{2}{(1+x^{2})} \frac{d}{dx}(\frac{1}{(1+t^{2})})

\frac{d^{2}y}{dx^{2}} = \frac{2}{(1+t^{2})}. \frac{-2x}{(1+x^{2})^{2}}+\frac{2}{(1+x^{2})}.\frac{-2t}{(1+t^{2})^{2}}.\frac{dt}{dx}

=\frac{-4x}{(1+t^{2})(1+x^{2})^{2}} - \frac{4t}{(1+x^{2})^{2}(1+t^{2})^{2}}

\frac{d^{2}y}{dx^{2}} = \frac{-4x(1+t^{2})-4t}{(1+x^{2})^{2}(1+t^{2})^{2}} = \frac{-4(t+x(1+t^{2}))}{(1+x^{2})^{2}(1+t^{2})^{2}} which in turn equals

\frac{-4(\tan{x}+x (1+\tan^{2}{x}))}{(1+x^{2})^{2}(1+\tan^{2}{x})^{2}} so where did we go wrong….quite clearly, practice alone can help us develop foresight…below is a cute proof:

\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{d}{dt}(\arctan{(\frac{2t}{1-t^{2}})})}{\frac{d}{dt}\arcsin(\frac{2t}{1+t^{2}})} and put t=\tan{\theta}

so that \frac{dy}{dx} = \frac{2\frac{d\theta}{dt}}{2\frac{d\theta}{dt}}=1 so we have bingo 🙂 an elegant answer



Nalin Pithwa.

Derivatives: part 12:IITJEE maths tutorial problems for practice

1, x = a(t+\frac{1}{t}), y=a(t-\frac{1}{t}), then find \frac{dy}{dx}.

Option (A) \frac{t^{2}-1}{t^{2}+1}

Option (B) \frac{t^{2}+1}{t^{2}-1}

Option (C) \frac{t^{2}+1}{1-t^{2}}

Option (D) \frac{1}{t}

Solution 1: Given that x=at+\frac{a}{t} so that \frac{dx}{dt} = a +\frac{a(-1)}{t^{2}} = a(1-\frac{1}{t^{2}})

and given that y = at - \frac{a}{t} so that \frac{dy}{dt} = x + \frac{a}{t^{2}} = a(1+ \frac{1}{t^{2}})

and so we get \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a(1 - \frac{1}{t^{2}})}{a(1+\frac{1}{t^{2}})} = \frac{t^{2}-1}{t^{2}+1} so that correct choice is option A.

2. If x = a \sin{3t} + b \cos{3t} and y= b \cos {t} + a \sin{t} then find \frac{dy}{dx} when t = \frac{\pi}{4}

Option (A) 0

Option (B) \frac{b-a}{3(a+b)}

Option (C) \frac{a-b}{3(a+b)}

Option (D) \frac{b-a}{b+a}

Solution 2:

\frac{dy}{dt} = -b \sin{t} + a\cos{t}

\frac{dx}{dt} = 3a \cos{3t} + -3b \sin{3t}

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = - \frac{-b\sin{t}+a \cos{t}}{3a \cos{3t}-3b\sin{3t}} which is equal to the following at t = \frac{\pi}{4}

\frac{dy}{dx} = \frac{- \frac{b}{\sqrt{2}} + \frac{a}{\sqrt{2}}}{-\frac{3a}{\sqrt{2}} - \frac{3b}{\sqrt{2}}}=-\frac{1}{3} \times \frac{b-a}{a+b} so that the correct choice is C.

3. If y = \frac{x \arcsin{x}}{\sqrt{1-x^{2}}} + \log{(\sqrt{1-x^{2}})}, then find \frac{dy}{dx}

Option A: \frac{\arcsin{x}}{(1-x^{2})^{\frac{3}{2}}}

Option B: \frac{\arcsin{x}}{\sqrt{1-x^{2}}}

Option C:\frac{\arcsin{x}}{1-x^{2}}

Option D: (1-x^{2})^{\frac{3}{2}}\arcsin{x}

Solution 3:

Let y = f(x) + g(x) where we put f(x) = \frac{x \arcsin{x}}{\sqrt{1-x^{2}}} so now let x=\sin{\theta}

So, we get \frac{dx}{d\theta} = \cos{\theta} and 1-x^{2}= \cos^{2}{\theta} and \sqrt{1-x^{2}} = \cos{\theta}

So we get f(\theta) = \frac{\theta \times \sin{\theta}}{\cos{\theta}} = \theta \times \tan{\theta}

So now \frac{df}{d\theta}= \tan{\theta}+ \theta \times \sin^{2}{\theta}

And, g(x) = \log{\sqrt{1-x^{2}}}

\frac{dg}{dx} =  \frac{1}{\sqrt{1-x^{2}}} \times \frac{d}{dx} (\sqrt{1-x^{2}}) = \frac{1}{2(1-x^{2})} \times (-2x)= \frac{-x}{1-x^{2}}

Hence, we get the following:

\frac{dy}{dx} = \frac{x}{\sqrt{1-x^{2}}}  + \frac{\arcsin{x}}{\frac{1}{1-x^{2}}} + \frac{-x}{1-x^{2}}

Question 4: Find the following: \frac{d}{dx}(sec^{-1}{(\frac{1}{\sqrt{1-x^{2}}})} + cot^{-1}(\frac{\sqrt{1-x^{2}}}{x}))

Option a: \frac{2}{\sqrt{1-x^{2}}}

Option b: \frac{1}{\sqrt{1-x^{2}}}

Option c: \frac{\sqrt{1-x^{2}}}{x}

Option d: \sqrt{1-x^{2}}

Solution 4:

Let f(x)=y_{1}=sec^{-1}(\frac{1}{\sqrt{1-x^{2}}})

Let x=\sin{\theta}, 1-x^{2}=\cos^{2}(\theta), \sqrt{1-x^{2}}=\cos{\theta}, and \frac{1}{\sqrt{1-x^{2}}} = \sec{\theta}

so y_{1}=\theta=\arcsin{x}

\frac{dy_{1}}{dx} = \frac{1}{\sqrt{1-x^{2}}}

Let y_{2}=cot^{-1}{\frac{\sqrt{1-x^{2}}}{x}}

Let x=\sin{\theta}, \sqrt{1-x^{2}}=\cos{\theta} and \frac{\sqrt{1-x^{2}}}{x}=\cot{\theta}

Let y_{2}= \cot^{-1}{\cot{\theta}}=\theta

so that \frac{dy_{2}}{dx} = \frac{d}{dx}(\arcsin{x})=\frac{1}{1-x^{2}}

so that \frac{dy}{dx}=\frac{2}{\sqrt{1-x^{2}}} so the option is a.

Question 5:

If y=(x+\sqrt{1+x^{2}})^{n} then find (x^{2}+1)(\frac{dy}{dx})^{2}.

Solution 5:


\frac{dy}{dx} = x(x+\sqrt{1+x^{2}})^{n-1}\frac{d}{dx}(x+\sqrt{1+x^{2}}) = n(x+\sqrt{1+x^{2}})^{n-1}(1+\frac{2x}{2\sqrt{1+x^{2}}})

(\frac{dy}{dx})^{2}.(x^{2}+1) = (x^{2}+1).n^{2}.(x+\sqrt{1+x^{2}})^{2n-2} \times (1+\frac{x}{\sqrt{1+x^{2}}})^{2}

=(x^{2}+1).n^{2}. (x+\sqrt{1+x^{2}})^{2n-2} \times (\frac{x+\sqrt{1+x^{2}}}{\sqrt{1+x^{2}}})^{2}


Question 6:

If f(x)= \sqrt{\frac{(x+1)(x+2)}{(x+3)(x+6)}}, then f^{'}(0) is equal to

(a) 1/2 (b) 1/3 (c) 1/6 (d) 0

Solution 6:

Given that y = \sqrt{\frac{(x+1)(x+2)}{(x+3)(x+6)}}

Hence, we have (x+3)(x+6) y^{2}=(x+1)(x+2)

(x+3)\frac{d}{dx}(y^{2}(x+6))+y^{2}(x+6) \times 1 = (x+1) \times 1 + (x+2) \times 1=2x+3

(x+3)(x+6)2y\frac{dy}{dx} + (x+3)y^{2} \times 1 =2x+3

(x+3)(x+6)2y\frac{dy}{dx} + (x+3) \times \frac{(x+1)(x+2)}{(x+3)(x+6)} = 2x+3

(x+3)(x+6)^{2}.2. \sqrt{\frac{(x+1)(x+2)}{(x+3)(x+6)}}+(x+1)(x+2) = (2x+3)(x+6)

2\sqrt{(x+1)(x+2)(x+3)(x+6)} \times (x+6)\frac{dy}{dx} + (x+1)(x+2) = (2x+3)(x+6)

So, at x=0, on substitution we get f^{'}(0).

Question 7:

If y = \frac{1-t^{2}}{1+t^{2}}, x = \frac{2t}{1+t^{2}}, then find \frac{dy}{dx}.

Solution 7:

Given y= \frac{1-t^{2}}{1+t^{2}}, let t= \tan{\theta} so that \frac{dt}{d\theta}= \sec^{2}(\theta)

so that y = \frac{1-t^{2}}{1+t^{2}} = \frac{1-\tan^{2}{\theta}}{1+\tan^{2}{\theta}} = \frac{\cos^{2}{\theta}-\sin^{2}{\theta}}{1} = 2 \cos^{2}{\theta}-1= \cos{2\theta}

Now, x = \frac{2t}{1+t^{2}}=\frac{2\tan{\theta}}{1+\tan^{2}{\theta}} so that x = \sin{2\theta}

so now \frac{dx}{d\theta}=2 \cos{2\theta}

y = \cos{2\theta}

\frac{dy}{d\theta} = -2 \sin{2\theta}

\frac{dy}{dx} = - \frac{2\sin{(2\theta)}}{2(\cos{2\theta})}= - \tan{(2\theta)} = - \frac{2t}{1-t^{2}}= - \frac{x}{y}.

Question 8:

Find \frac{d}{dx}(\arctan{x} + \arcsin{(\frac{x}{\sqrt{1+x^{2}}})})

Solution 8:

Let it be given that y = \arctan{x} + \arcsin{(\frac{x}{\sqrt{1+x^{2}}})}

Now, let us simplify this as y=y_{1}+y_{2} where y_{1} = \arctan{x} and y_{2} = \arcsin{(\frac{x}{\sqrt{1+x^{2}}})}

Now, first consider y_{1} = \arctan{x}. Taking derivative of both sides w.r.t. x, we get

\frac{dy_{1}}{dx} = \frac{d}{dx}(\arctan{x}) = \frac{1}{1+x^{2}}….A

Now, next consider y_{2} = \arcsin{(\frac{x}{\sqrt{1+x^{2}}})}. Takind derivative of both sides w.r.t. x, we get

\frac{dy_{2}}{dx} = \frac{d}{dx}(\arcsin{(\frac{x}{\sqrt{1+x^{2}}})}) = \frac{1}{1- \frac{x^{2}}{1+x^{2}}} \frac{d}{dx}(\frac{x}{\sqrt{1+x^{2}}}) = \frac{1}{1+x^{2}}(\frac{1}{\sqrt{1+x^{2}}+\frac{x}{2}(1+x^{2})^{-3/2}})….B

So that we get \frac{dy}{dx} = \frac{dy_{1}}{dx} + \frac{dy_{2}}{dx}using A and B.

Question 9:

If x^{y} = y^{x}, then find \frac{dy}{dx}

Solution 9:

Given that x^{y} = y^{x}

y \log{x}= x \log{y}. Taking derivative of both sides w.r.t. x, we get

(\log{x}).\frac{dy}{dx} + \frac{y}{x} = \frac{x}{y}. \frac{dy}{dx} + (\log{y}) \times 1

(\log{x}- \frac{x}{y}).\frac{dy}{dx} = (\log{y}) - \frac{y}{x}

\frac{dy}{dx} = \frac{\frac{x(\log{y}-y)}{x}}{\frac{y\log{x}-x}{y}}= \frac{y}{x} \times \frac{x(\log{y})-y}{y(\log{x})-x} which is the required answer.

Question 10:

If (x+y)^{m+n} = x^{m}y^{n}, then find \frac{dy}{dx}.

Solution 10:

Given that (x+y)^{m+n} = x^{m}y^{n}

Taking logarithm of both sides w.r.t. any arbitrary valid base,

(m+n) \times \log{(x+y)} = \log{(x^{m}y^{n})} = \log(x^{m}) + \log{y^{n}} so that (m+n).\log{(x+y)}=m \log{x} + n \log{y}

Taking derivative of both sides w.r.t. x, we get the following:

\frac{m+n}{x+y}. \times (1+\frac{dy}{dx}) = \frac{m}{x} + \frac{n}{y}.\times \frac{dy}{dx}

\frac{m+n}{x+y} \times \frac{dy}{dx} - \frac{x}{y}. \frac{dy}{dx} = \frac{m}{x} - \frac{m+n}{x+y}

\frac{(m+n)y-n(x+y)}{y(x+y)}. \frac{}{} = \frac{mx+my-mx-nx}{x(x+y)}

\frac{my+ny-nx-ny}{} = \frac{my-nx}{x(x+y)}

\frac{my-nx}{y(n+y)}. \frac{dy}{dx} = \frac{my-nx}{x(x+y)}, so that finally we get the desired answer:

\frac{dy}{dx} = \frac{y}{x}

More later,


Nalin Pithwa

Derivatives: part 11: IITJEE maths tutorial problems for practice

Problem 1: Find \frac{d}{dx}\arctan{\frac{4x}{4-x^{2}}}.

Choose (a) \frac{1}{4-x^{2}} (b) \frac{1}{4+x^{2}} (c) \frac{4}{4+x^{2}} (d) \frac{4}{4-x^{2}}

Solution 1:

Let y = \arctan{\frac{4x}{4-x^{2}}}. Hence, \tan{y} = \frac{4x}{4-x^{2}}. Differentiating both sides w.r.t. x, we get the following:

\sec^{2}{y} \times \frac{dy}{dx}= \frac{d}{dx} (\frac{4x}{4-x^{2}})

\sec^{2}{y} \times \frac{dy}{dx} = \frac{(4-x^{2}) \times 4 - 4x \times (-2x)}{(4-x^{2})^{2}} = \frac{16+4x^{2}}{(4-x^{2})}

But, \sec^{2}{y}=\tan^{2}{y}+1=\frac{(x^{2}+4)^{2}}{(4-x^{2})^{2}}

Hence, the answer is \frac{dy}{dx}= \frac{4}{4+x^{2}}. Option c.

Problem 2: Find \frac{dy}{dx} if \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}}=1

Choose (a) \frac{-(2x+y)}{x+y} (b) \frac{-(2x+y)}{x+2y} (c) \frac{x+2y}{x+y} (d) -\frac{2x+y}{x+2y}

Solution 2:

The given equation is x+y = \sqrt{xy}. Differentiating both sides wrt x,

1+ \frac{dy}{dx} = \sqrt{y} \times \frac{d}{dx} (\sqrt{x})+ \sqrt{x}\frac{d}{dx}(y^{\frac{1}{2}})

1+\frac{dy}{dx} = \frac{\sqrt{y}}{2\sqrt{x}} + \frac{\sqrt{x}}{2\sqrt{y}} \times \frac{dy}{dx}

(1- \frac{1}{2}\sqrt{\frac{x}{y}}) \times \frac{dy}{dx} =\frac{\sqrt{y}}{2\sqrt{x}} -1

\frac{dy}{dx} = \frac{\sqrt{y}-2\sqrt{x}}{2\sqrt{y}-\sqrt{x}} \times \frac{2\sqrt{y}}{2\sqrt{x}}

\frac{}{} = \frac{2y - 4\sqrt{xy}}{4\sqrt{xy}-2x} = \frac{2y-4(x+y)}{4(x+y)-2x} = - \frac{2x+y}{x+2y} is the answer. Option D.

Problem 3: If y=\arctan{\frac{\log{(\frac{e}{x^{2}})}}{\log{(ex^{2})}}} then \frac{dy}{dx} is

choose (a) e (b) \frac{2}{x(1+4(\log{x})^{2})}(c) \frac{-2}{x(1+4(\log{x})^{2})} (d) \frac{2}{1+x^{2}}

Solution 3:

Given that y = \arctan{(\frac{\log(\frac{e}{x^{2}})}{\log(ex^{2})})} so that we have

\tan{y} = \frac{\log{\frac{e}{x^{2}}}}{\log{ex^{2}}} so now differentiating both sides w.r.t. x,

\sec^{2}{y}\frac{dy}{dx} = \frac{\frac{\log{(ex^{2})}}{\frac{e}{x^{2}}} \frac{d}{dx}(\frac{e}{x^{2}}) - \log{(\frac{e}{x^{2}})} \times \frac{1}{ex^{2}} \times \frac{d}{dx}(ex^{2})}{(\log{(ex^{2})})}

\sec^{2}{y}\frac{dy}{dx} = \frac{-\frac{2}{x}(\log{(ex^{2})})- \frac{2}{x}\log{(\frac{e}{x^{2}})}}{(\log{(ex^{2})})^{2}}

\sec^{y}(\frac{dy}{dx}) = \frac{-\frac{2}{x}(\log{(ex^{2}) \times (\frac{e}{x^{2}})})}{(\log{ex^{2}})^{2}}

\sec^{y}\frac{dy}{dx} = \frac{-\frac{4}{x}}{(\log{(ex^{2})^{2}})} = \frac{-4}{x(\log{(ex^{2})})^{2}}

Now, we also know that\sec^{2}{y} = 1 + \tan^{2}{y} = \frac{(\log{(\frac{e}{x^{2}})})^{2}}{(\log{(ex^{2})})^{2}} + 1 = \frac{(\log(\frac{e}{x^{2}}))^{2}+(\log{(ex^{2})})^{2}}{(\log{(ex^{2})})^{2}}

But, note that by laws of logarithms, on simplification, we get

\log{(\frac{e}{x^{2}})} = 1 - 2\log{x} and \log{(ex^{2})} = 1 + 2 \log{x} so that on squaring, we get

(\log{(e/x^{2})})^{2} = 1-4\log{x} + 4 (\log{x})^{2}

(\log{(ex^{2})})^{2}=1+4\log{x} + 4 (\log{x})^{2} so that now we get

(\log{(\frac{e}{x^{2}})})^{2} + (\log{(ex^{2})})^{2} = 2 + 8 (\log{x})^{2}, which all put together simplifies to

\frac{dy}{dx} = \frac{1}{\sec^{2}{y}} \times \frac{-(\frac{4}{x})}{(\log{(ex^{2})})^{2}}

\frac{dy}{dx} = - \frac{(\frac{2}{x})}{1+4(\log{x})^{2}} so that the answer is option C.

Problem 4: Find \frac{d}{dx}(\arcsin{(3x-4x^{3})}+\arccos{(2x(\sqrt{(1-x^{2})}))})

Choose option (a) \frac{1}{\sqrt{1-x^{2}}} (b) \frac{-1}{\sqrt{1-x^{2}}} (c) \frac{5}{\sqrt{1-x^{2}}} (d) \frac{-2}{\sqrt{1-x^{2}}}

Solution 4:

Let us consider the first differential. Let us substitute x = \sin{\theta}. Hence,

3x-4x^{3}=3\sin{\theta} - 4\sin^{3}{\theta}= \sin{3\theta} and so we \arcsin{3x-4x^{3}} = \arcsin{\sin{3\theta}} = 3 \theta, and so also, we get \arccos{2x\sqrt{1-x^{2}}}=\arccos{2\sin{\theta}\cos{\theta}} = \arccos{\cos{2\theta}}=2\theta so we get

required derivative

\frac{dy}{dx} = \frac{d}{dx}(3\theta) + \frac{d}{dx}(2\theta) = \frac{d}{dx}(5\theta) = 5 \frac{d\theta}{dx} = 5 \frac{d}{dx}(\arcsin{x})= 5 \frac{1}{\sqrt{1-x^{2}}}. Answer is option C.

Problem 5: Find \frac{d}{dx}(x-a)(x-b)(x-c)\ldots (x-z)

Choose option (a) zero (b) 26 (c) 26! (d) does not exist

Solution 5: the expression also includes a term 0 = (x-x) so that the final answer is zero only.

Problem 6: Find \frac{d}{dx}(x^{x})^{x}.

Solution 6: Let y= (x^{x})^{x}

so \log{y} = \log{(x^{x})^{x}}

so \log{y} = x^{2} \times \log{x} so that differentiating both sides w.r.t. x, we get

we get \frac{1}{y} \times \frac{dy}{dx} = \frac{x^{2}}{x} + \log{x} \times 2x

we get \frac{1}{y} \times \frac{dy}{dx} = x + 2x \log{x} = x(1+2\log{x})

we get \frac{dy}{dx} = yx (1+2 \log{x}) = (x^{x})^{x} \times x \times (1+2\log{x})

so the answer is option B.

Choose option (a): x.x^{x}(1+2\log{x}) (b) x^{x^{2}+1} \times (1+2\log{x}) (c) {x^{{x}^{2}}}(1+\log{x}) (d) none of these

Problem 7:

Find \frac{d}{dx}(e^{x^{x}})

Choose option (a) e^{x^{x}}.x^{x}.(1+\log{x}) (b) e^{x^{x}}. x^{x}.\log{(\frac{x}{e})} (c) e^{x^{x}}.x^{x} (d) e^{x^{x}}. (\log{(e^{x})})

Solution 7: Let y = (e^{(x^{x})}) so that taking logarithm of both sides

\log{y} = \log{(e^{(x^{x})})} so that \log{y} = x^{x} \log{e} = x^{x}

\log {(\log{y})}= x \times (\log{x}). Differentiating both sides w.r.t.x we get:

\frac{1}{\log{y}} \times \frac{d}{dx} \times (\log{y})= \frac{x}{x} + \log{x} so that we get now

$latex\frac{1}{y(\log{y})} \times \frac{dy}{dx} = 1 + \log{x} $

\frac{dy}{dx} = e^{x^{x}} \times x^{x} \times (1+\log{x}) so we get option a as the answer.

Problem 8:

Find \frac{d}{dx}(x^{x^{x}})

Choose option (a): x^{x^{x}} \times (1+\log{x}) (b) x^{x^{x}} \times (x^{x}\log{x})(1+\log{x}+\frac{1}{x}) (c) x^{x^{x}} \times (x^{x}\log{x}) \times (1+\log{x}+\frac{1}{x\log{x}}) (d) none of these.

Solution 8:

let y=x^{x^{x}} taking logarithm of both sides we get

\log{y} = x^{x} \times \log{x} and now differentiating both sides w.r.t.x, we get

\frac{1}{y} \times \frac{dy}{dx} = \frac{x^{x}}{x} + (\log{x}) \times \frac{d}{dx} (x^{x}) and now let t=x^{x} and again take logarithm of both sides so that we get (this is quite a classic example…worth memorizing and applying wherever it arises):

\log{t}= x\log{x}

\frac{1}{t} \frac{dt}{dx} = \frac{x}{x} + \log{x}=1+\log{x}

\frac{dt}{dx} = x^{x}(1+\log{x})

\frac{dy}{dx} \times \frac{1}{y} = x^{x-1} + (\log{x}).x^{x}.(1+\log{x})

\frac{dy}{dx} = x^{x}(x^{x-1}+x^{x} \times \log{x} \times (1+\log{x}))

\frac{dy}{dx} = x^{x^{x}} (x^{x} \times (\log{x})) \times (1+ \log{x}+ \frac{1}{x \log{x}})

The answer is option C.

Problem 9:

Find \frac{d}{dx}(x+a)(x^{2}+a^{2})(x^{4}+a^{4})(x^{8}+a^{8}).

Choose option (a): \frac{15x^{16}-16x^{15}a+a^{16}}{(x-a)^{2}} (b) \frac{x^{16}-a^{16}}{x-a} (c) \frac{x^{16}-x^{15}a+a^{16}}{(x-a)^{2}} (d) none of these

Solution 9:

Given that y = (x+a)(x^{2}+a^{2})(x^{4}+a^{4})(x^{8}+a^{8})

Remark: Simply multpilying out thinking the symmetry will simplify itself is going to lead to a mess…because there will be no cancellation of terms …:-) The way out is a simple algebra observation…this is why we should never ever forget the fundamentals of our foundation math:-)

note that the above can be re written as follows:

y = \frac{(x^{2}-a^{2})}{(x-a)} \times \frac{(x^{4}-a^{4})}{(x^{2}-a^{2})} \times \frac{x^{8}-a^{8}}{(x^{4}-a^{4})} \times \frac{(x^{16}-a^{16})}{(x^{8}+a^{8})}

Now, we are happy like little children because many terms cancel out 🙂 hahaha…lol 🙂

y = \frac{(x^{16}-a^{16})}{(x-a)} and now differentiating both sides w.r.t.x we get

\frac{dy}{dx} = \frac{(x-a)(16x^{15})- (x^{16}-a^{16})(1)}{(x-a)^{2}}

\frac{dy}{dx} = \frac{15x^{16}-16x^{15}a+a^{16}}{(x-a)^{2}}

The answer is option A.

Problem 10:

If x= \theta {\cos{\theta}}+\sin{\theta} and y = \cos{\theta}-\theta \times \sin{\theta} then find the value of \frac{dy}{dx} at\theta = \frac{\pi}{2}

Choose option (a): -\frac{\pi}{2} (b) \frac{2}{\pi} (c) \frac{\pi}{4} (d) \frac{4}{\pi}

Solution 10:

\frac{dx}{d\theta} = \cos{\theta} - \theta \times \sin{\theta} + \cos{\theta}

\frac{dy}{d\theta} = -\sin{\theta} - (\sin{\theta} + \theta \times \cos{\theta})

\frac{dy}{d\theta} = \theta \times \cos{\theta} - 2\sin{\theta}

\frac{dx}{d\theta} = 2 \cos{\theta} - \theta \times \sin{\theta}

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dy}{d\theta}} = \frac{\theta \times \cos{\theta}-2\sin{\theta}}{2\cos{\theta}-\theta \times \sin{\theta}} = \frac{-2}{-\pi/2}=\frac{4}{\pi}

Answer is option D.


Nalin Pithwa.

A problem of log, GP and HP…

Question: If a^{x}=b^{y}=c^{z} and b^{2}=ac, pyrove that: y = \frac{2xz}{x+z}

Solution: This is same as proving: y is Harmonic Mean (HM) of x and z;

That is, to prove that y=\frac{2xz}{x+z} is the same as the proof for : \frac{1}{y} - \frac{1}{x} = \frac{1}{z} - \frac{1}{y}

Now, it is given that a^{x} = b^{y} = c^{z} —– I

and b^{2}=ac —– II
Let a^{x} = b^{y}=c^{z}=N say. By definition of logarithm,

x = \log_{a}{N}; y=\log_{b}{N}; z=\log_{c}{N}

\frac{1}{x} = \frac{1}{\log_{a}{N}}; \frac{1}{y} = \frac{1}{\log_{b}{N}}; \frac{1}{z} = \frac{1}{\log_{a}{N}}.

Now let us see what happens to the following two algebraic entities, namely, \frac{1}{y} - \frac{1}{x} and \frac{1}{z} - \frac{1}{y};

Now, \frac{1}{y} - \frac{1}{x} = \frac{1}{\log_{b}{N}} - \frac{1}{\log_{a}{N}} = \frac{\log_{b}{b}}{\log_{b}{N}} - \frac{\log_{a}{a}}{\log_{a}{N}} = \log_{N}{b} - \log_{N}{a} = \log_{N}{(\frac{b}{a})}…call this III

Now, \frac{1}{z} - \frac{1}{y} = \frac{1}{\log_{c}{N}} - \frac{1}{\log_{b}{N}} = \frac{\log_{c}{c}}{\log_{c}{}N} -\frac{\log_{b}{b}}{\log_{b}{N}}= \log_{N}{c}-\log_{N}{b}

Hence, \frac{1}{z} - \frac{1}{y}=\log_{N}{c/b}….equation IV

but it is also given that b^{2}=ac…see equation II

Hence, \frac{b}{a} = \frac{c}{b}

Take log of above both sides w.r.t. base N:

So, above is equivalent to \log_{N}{b/a} = \log_{N}{c/b}

But now see relations III and IV:

Hence, \frac{1}{y} -\frac{1}{x} = \frac{1}{x} - \frac{1}{y}

Hence, \frac{2}{y} = \frac{1}{x} + \frac{1}{z} = \frac{x+z}{xz}

Hence, y= \frac{2xz}{x+z} as desired.


Nalin Pithwa

Derivatives: Part 10: IITJEE maths tutorial problems for practice

Problem 1: If x=3\cos{\theta}-\cos^{3}{\theta}, and y=3\sin{\theta}-\sin^{3}{\theta}, then \frac{dy}{dx} is equal to:

(a) -\cot^{3}{\theta} (b) -\tan^{3}{\theta} (c) \cot^{3}{\theta} (d) \tan^{3}{\theta}

Problem 2: If x = \tan{\theta} + \cot{\theta}, and y=2 \log{(\cot{\theta})}, then \frac{dy}{dx} is equal to:

(a) \tan{(2\theta)} (b) \cot{(2\theta)} (c) \tan{\theta} (d) \sec^{2}{2\theta}

Problem 3: \frac{d}{dx}\log{\sqrt{\frac{1-\cos{x}}{1+\cos{x}}}} is equal to:

(a) \tan{\frac{x}{2}} (b) \sin{x} (c) cosec(x) (d) \tan{x}

Problem 4: y=2^{2(\log_{2}{(x+2)}-\log_{2}{(x+1)})}, then \frac{dy}{dx} is:

(a) \frac{-2(x+2)}{(x+1)^{3}} (b) \frac{4(x+2)}{(x+1)^{3}} (c) \frac{2(x+2)}{(x+1)^{3}} (d) \frac{-6(x+2)}{(x+1)^{3}}

Problem 5: \frac{d}{dx}(\arctan{\sqrt{\frac{e^{x}-1}{e^{x}+1}}}) is equal to:

(a) \frac{1}{1+e^{2x}} (b) \frac{1}{2\sqrt{e^{2x}-1}} (c) \frac{e^{x}}{2\sqrt{1+e^{2x}}} (d) \frac{1}{2\sqrt{1-e^{2x}}}

Problem 6: y=e^{m\arcsin{x}} then (1-x^{2}) (y^{'})^{2} is equal to :

(a) y^{2} (b) m^{2}(1-y^{2}) (c) -m^{2}y^{2} (d) m^{2}y^{2}

Problem 7: If y = \sin(m \arcsin{x}) then (1-x^{2})(\frac{dy}{dx})^{2} is

(a) m^{2}y^{2} (b) m^{2}(1-y^{2}) (c) -m^{2}y^{2} (d) m^{2}(1+y^{2})

Problem 8: \frac{d}{dx}(\cos{\arctan{x}}) is:

(a) \frac{1}{2\sqrt{1+x^{2}}} (b) \frac{-x}{(1+x^{2})^{\frac{3}{2}}}

(c) \frac{-x}{\sqrt{1+x^{2}}} (d) \frac{2x}{\sqrt{1+x^{2}}}

Problem 9: If y = \arcsin{\frac{a\cos{x}+b\sin{x}}{\sqrt{a^{2}+b^{2}}}} then \frac{d^{2}y}{dx^{2}} is:

(a) -1 (b) 0 (c) 1 (d) \arctan{\frac{b}{a}}

Problem 10: \arctan{\frac{4x}{4-x^{2}}} is:

(a) \frac{1}{4-x^{2}} (b) \frac{1}{4+x^{2}} (c) \frac{4}{4+x^{2}} (d) \frac{4}{4-x^{2}}


Nalin Pithwa.

Derivatives: Part 9: IITJEE maths tutorial problems practice

Problem 1: \frac{d}{dx}((\frac{1}{b}\arctan{\frac{x}{b}})-\frac{1}{a}\arctan{(\frac{x}{a})}) is equal to:

(a) \frac{1}{(x^{2}+a^{2})(x^{2}+b^{2})} (b) \frac{a^{2}-b^{2}}{(x^{2}+a^{2})(x^{2}+b^{2})}

(c) \frac{x^{2}+a^{2}}{x^{2}+b^{2}} (d) \frac{2x^{2}}{(x^{2}+a^{2})(x^{2}+b^{2})}

Problem 2: \frac{d}{dx}(\frac{x}{2} + \frac{1}{2}\log{(\sin{x}+\cos{x})}) is equal to:

(a) \frac{\tan{x}}{1+\tan{x}} (b) \frac{1}{1+\cot{x}} (c) \frac{1-\tan{x}}{1+\tan{x}} (d) \frac{1}{1+\tan{x}}

Problem 3: If y=\sqrt{\frac{cosec{x}-\cot{x}}{cosec{x}+\cot{x}}} where 0<x<\frac{\pi}{2}, then \frac{dy}{dx} is given by :

(a) cosec{x}(cosec{x}-\cot{x}) (b) cosec{x}(\cot{x}-cosec{x}) (c) cosec{x}(\cot{x}-cosec{x}) (d) \cot{x}(cosec{x}-\cot{x})

Problem 4: \frac{d}{dx}\log {|\sec{(x-\frac{\pi}{4})}+\tan{(x-\frac{\pi}{4})}|} is equal to:

(a) \frac{\sqrt{2}}{\sin{x}-\cos{x}} (b) \frac{\sin{x}}{\sin{x}+\cos{x}} (c) \frac{\sqrt{2}}{\sin{x}+\cos{x}} (d) \frac{1}{\sin{x}+\cos{x}}

Problem 5:

If r=a(1+\cos{\theta}), and \tan{\phi}=r\frac{d\theta}{dr}, then \phi is equal to:

(a) \frac{-2}{\theta} (b) \frac{\pi}{2} + \frac{\theta}{2} (c) -\frac{\theta}{2} (d) \frac{\pi}{2} - \frac{\theta}{2}

Problem 6: \frac{d}{dx}\log{(\sqrt{x+ \sqrt{x^{2}+a^{2}}})} is equal to:

(a) \frac{1}{2\sqrt{x^{2}+a^{2}}} (b) \frac{1}{x+\sqrt{x^{2}+a^{2}}} (c) \frac{1}{\sqrt{x^{2}+a^{2}}} (d) \frac{1}{2(x+\sqrt{x^{2}+a^{2}})}

Problem 7: \frac{d}{dx}(\log{(1+\sin{(2x)})} + 2 \log{\sec{(\frac{\pi}{4}-x)}}) is equal to

(a) 0 (b) \log{2} (c) \frac{4(\cos{x}-\tan{x})}{\sin{x}+\cos{x}} (d) \frac{2\cos{(2x)}}{1+\sin{(2x)}} + \tan{(\frac{\pi}{4}-x)}

Problem 8: If x^{2}+xy+y^{2}=1, then \frac{dy}{dx} is equal to:

(a) -\frac{x+2y}{y+2x} (b) -\frac{y+2x}{x+2y} (c) \frac{y+2x}{x+2y} (d) \frac{2(x+y)}{y-2x}

Problem 9: \frac{d}{dx}(\arcsin{(\sqrt{\frac{1-x}{2}})}) is equal to:

(a) \frac{1}{\sqrt{1-x^{2}}} (b) \frac{-1}{\sqrt{1-x^{2}}} (c) \frac{1}{2\sqrt{1-x^{2}}} 9d) \frac{-1}{2\sqrt{1-x^{2}}}

Problem 10: If y = \arctan{(\frac{3a^{2}x-x^{3}}{x^{3}-3ax^{2}})} then \frac{dy}{dx} is equal to:

(a) \frac{3}{a} (b) \frac{1}{a} (c) \frac{3x}{a} (d) \frac{3a}{x^{2}+a^{2}}


Nalin Pithwa

Derivatives: part 8: IITJEE mains tutorial problems practice

Problem 1: If y=b(\arctan{(\frac{x}{y})})+ \arctan{(\frac{y}{x})}, then \frac{dy}{dx} is equal to:

(a) \frac{x}{2} (b) -1 (c) 0 (d) b

Problem 2: If r=a(1+\cos{\theta}), then \sqrt{r^{2}+(\frac{dr}{d\theta})^{2}} is:

(a) 2a\cos{\theta} (b) 2a \sin{(\frac{\theta}{2})} (c) 2a \cos{(\frac{\theta}{2})} (d) 2a \sin{\theta}

Problem 3: \frac{d}{dx}\arctan{\log_{10}{x}} is equal to:

(a) \frac{1}{1 + (\log_{10}{x})^{2}} (b) \frac{1}{x \log_{10}{(1+ (\log_{10}{x})^{2})}} (c) \frac{1}{x(1+(\log_{10}{x})^{2})} (d) \frac{1}{10 \log{x}(1+(\log_{10}{x})^{2})}

Problem 4: If \sin^{2}(mx) + \cos^{2}(ny)=a^{2}, then \frac{dy}{dx} is equal to:

(a) \frac{m \sin{(2mx)}}{n \sin{(2ny)}} (b) \frac{n\sin{(2mx)}}{m\sin{(2ny)}} (c) \frac{n\sin{(2ny)}}{m\sin{(2mx)}} (d) \frac{-m\sin{(2mx)}}{n\sin{(2ny)}}

Problem 5: \frac{d}{dx}(\frac{\tan{x}-\cot{x}}{\tan{x}+\cot{x}}) is equal to:

(a) 2\sin{(2x)} (b) \sin{(2x)} (c) -2 \sin{(2x)} (d) 2\cos{(2x)}

Problem 6: If y=\log_{5}{(\log_{5}{x})} then the value of \frac{dy}{dx} is

(a) \frac{1}{x \log_{5}{x}} (b) \frac{1}{x \log_{5}{x}. (\log{5})^{2}} (c) \frac{1}{\log{5}.x\log{x}} (d) \frac{1}{x(\log_{5}{x})^{2}}

Problem 7: \frac{d}{dx}(ax+b)^{cx+d} is equal to:

(a) (ax+b)^{cx+d}(\frac{cx+d}{ax+b} + \log{(ax+b)}) (b) (ax+b)^{cx+d}(\frac{cx+d}{ax+b} + c \log{(ax+b)}) (c) a(ax+b)^{cx+d} (d) none

Problem 8: \frac{d}{dx}(\log{(\frac{\sin{(x-b)}}{\sin{(x-a)}})}) is equal to:

(a) \frac{\cos{(a-b)}}{\sin{(x-a)}\sin{(x-b)}} (b) \frac{\sin{(b-a)}}{\sin{(x-a)}\sin{(x-b)}}

(c) \frac{\sin{(a-b)}}{\sin{(x-a)}\sin{(x-b)}} (d) \frac{\cos{(b-a)}}{\sin{(x-a)}\sin{(x-b)}}

Problem 9: If y = \sqrt{\frac{\sec{x}+\tan{x}}{\sec{x}-\tan{x}}} and 0<x<\frac{\pi}{2}, then \frac{dy}{dx} is :

(a) \sec{x}(\sec{x}-\tan{x}) (b) \sec{x}(\sec{x}+\tan{x}) (c) \tan{x}(\sec{x}+\tan{x}) (d) \tan{x}(\sec{x}-\tan{x})

Problem 10: \frac{d}{dx}e^{ax}(a\sin{(bx)}-b\cos{(bx)}) is equal to:

(a) e^{ax}(\sin{(bx)}) (b) (a^{2}+b^{2})e^{ax}\sin{(bx)} (c) e^{ax}\cos{(bx)} (d) (a^{2}+b^{2})e^{ax}\cos{(bx)}


Nalin Pithwa.

Derivatives: part 7: IITJEE tutorial problems practice

Problem 1: Differential coefficient of \sec{\arctan{x}} is

(a) \frac{x}{1+x^{2}} (b) x\sqrt{1+x^{2}} (c) \frac{1}{\sqrt{1+x^{2}}} (d) \frac{x}{\sqrt{1+x^{2}}}

Problem 2: If \sin{(x+y)} = \log{(x+y)}, then \frac{dy}{dx} is equal to :

(a) 2 (b) -2 (c) 1 (d) -1

Problem 3: If y = \arcsin{\sqrt{x-ax}-\sqrt{a-ax}}, then \frac{dy}{dx} is equal to:

(a) \frac{1}{2\sqrt{x}\sqrt{1-x}} (b) \sin{(\sqrt{x})} \times \sin{(\sqrt{a})}

(c) \frac{1}{\sin{\sqrt{a-ax}}} (d) zero

Problem 4: For the differentiable function f, the value of : \lim_{h \rightarrow 0} \frac{(f(x+h))^{2}-(f(x))^{2}}{2h} is equal to:

(a) (f^{'}(x))^{2} (b) \frac{1}{2}(f(x))^{2} (c) f(x)f^{'}(x) (d) zero

Problem 5: The derivative of \arctan{\frac{\sqrt{1+x^{2}}-1}{x}} w.r.t. \arctan{(\frac{2x\sqrt{1-x^{2}}}{1-2x^{2}})} at x=0 is :

(a) \frac{1}{8} (b) \frac{1}{4} (c) \frac{1}{2} (d) 1

Problem 6: If x = e^{y+e^{y+e^{y+e^{y+ \ldots}}}} then \frac{dy}{dx} is

(a) \frac{x}{1+x} (b) \frac{1}{x} (c) \frac{1-x}{x} (d) \frac{-1}{x^{2}}

Problem 7: Consider the following statements:

(1) (\frac{f}{g})^{'} = \frac{f^{'}}{g^{'}} (2) \frac{(fg)^{'}}{fg} = \frac{f^{'}}{f} + \frac{g^{'}}{g}

(3) \frac{(f+g)^{'}}{f+g} = \frac{f^{'}}{f} + \frac{g^{'}}{g} (4) \frac{(f/g)^{'}}{f/g} = \frac{f^{'}}{f} + \frac{g^{'}}{g}

Which of the following statements are true?

(a) 1 and 2 (b) 2 and 3 (c) 2 and 4 (d) 3 and 4

Problem 8: If y=e^{x+3\log{x}} then \frac{dy}{dx} =

(a) e^{x+3\log{x}} (b) e^{x}.x^{2}(x+3) (c) e^{x}. e^{3\log{x}} (d) 3x^{2}e^{x}

Problem 9: If y=\sin^{2}(x \deg), then find the value of \frac{dy}{dx} is:

(a) \frac{\pi}{360}\sin{(2 x \deg)} (b) \frac{\pi}{2}\sin{(2x\deg)} (c) 180 \sin {(2x\deg)} (d) \frac{\pi}{180}\sin{(2x\deg)}

Problem 10: If y=\log_{a}{x} + \log_{x}{a} + \log_{x}{x}+ \log_{a}{a} then the value of \frac{dy}{dx} is:

(a) \frac{1}{x}+x\log{a} (b) \frac{\log{a}}{x} + \frac{x}{\log{a}} (c) \frac{1}{x \log{a}}+ x \log{a} (d) \frac{1}{x\log{x}} - \frac{\log{a}}{x(\log{x})^{2}}


Nalin Pithwa.

Derivatives: part 6: IITJEE tutorial practice problems

Problem 1:

If \sec {(\frac{x+y}{x-y})}=a, then \frac{dy}{dx} is (i) \frac{x}{y} (ii) \frac{y}{x} (iii) y (iv) x

Problem 2:

If f(x) = x+ 2, when -1<x<1;

f(x)=5, when x=3;

f(x) = 8-x, when x>3; then, at x=3, the value of f^{'}(x) is

(a) 1 (b) -1 (c) 0 (d) does not exist.

Problem 3:

If y = x \tan{y}, then \frac{dy}{dx} is equal to

(i) \frac{\tan{y}}{x-x^{2}-y^{2}} (ii) \frac{\tan{y}}{y-x}

(iii) \frac{y}{x-x^{2}-y^{2}} (iv) \frac{\tan{x}}{x-y^{2}}

Problem 4:

If g is the inverse function of f and f^{'}(x) = \frac{1}{1+x^{n}}, then g^{'}(x) is equal to

(i) 1 + (g(x))^{n} (ii) 1+g(x) (iii) 1-g(x) (iv) 1-(g(x))^{n}

Problem 5:

If f(x) = \log_{x^{2}}(\log{x}) then f(x) at x=c is :

(i) 0 (ii) 1 (iii) \frac{1}{e} (iv) \frac{1}{2e}

Problem 6:

If y = (\sin{x})^{\tan{x}} then \frac{dy}{dx} is equal to :

(i) (\sin{x})^{\tan{x}}(1+ \sec^{2}{x} \log{\sin{x}})

(ii) \tan{x}. (\sin{x})^{\tan{x}-1} \times \cos{x}

(iii) (\sin{x})^{\tan{x}}\sec^{2}{x} \times \log{\sin{x}}

(iv) \tan{x} (\sin{x})^{\tan{x}-1}

Problem 7:

If y = \sqrt{\sin{x}+y}, then \frac{dy}{dx} equals:

(i) \frac{\sin{x}}{2y-1} (ii) \frac{\sin{x}}{1-2y} (iii) \frac{\cos{x}}{1-2y}

(iv) \frac{\cos{x}}{2y-1}

Problem 8:

If x = \sqrt{\frac{1-t^{2}}{1+t^{2}}} and y = \sqrt{\frac{\sqrt{1+t^{2}}-sqrt{1-t^{2}}}{\sqrt{1+t^{2}}+\sqrt{1-t^{2}}}}

then the value of \frac{d^{2}y}{dx^{2}} at t=0 is given by:

(a) 0 (b) 1/2 (c) 1 (d) -1

Problem 9:

If x = a \cos^{3}{\theta}, y = a \sin^{3}{\theta}, then \sqrt{1 + (\frac{dy}{dx})^{2}} is equal to:

(i) \sec^{2}{\theta} (ii) \tan^{2}{\theta} (iii) \sec{\theta} (iv) |\sec{\theta}|

Problem 10:

If y = \arcsin{\sqrt{1-x}} + \arccos{\sqrt{x}}, then \frac{dy}{dx} equals:

(a) \frac{1}{\sqrt{x(1-x)}} (b) \frac{1}{x(1+x)} (c) \frac{-1}{\sqrt{x(1-x)}} (d) none


Nalin Pithwa