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^{2}-y^{2}=y^{2}x-x^{2}y

(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

\frac{2\cos{x}}{(1-\sin{x})^{2}}

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

y=\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

\frac{d^{2}y}{dx^{2}}=1

Cheers,

Nalin Pithwa.

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