## 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 $\frac{dy}{dx}$.

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

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 $\frac{d^{2}y}{dx^{2}}$.

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.

This site uses Akismet to reduce spam. Learn how your comment data is processed.