Monthly Archives: April 2021

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.