Exercise 5.2

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Q.1 Differentiate $y=\sin(x^2+5)$ with respect to $x$.

Solution:

Given,

$$ y=\sin(x^2+5). $$

Using the chain rule,

$$ \frac{d}{dx}(\sin u)=\cos u\cdot\frac{du}{dx}, $$

where

$$ u=x^2+5. $$

Now,

$$ \frac{du}{dx}=2x. $$

Therefore,

$$ \begin{aligned} \frac{dy}{dx} &=\cos(x^2+5)\cdot2x\\ &=2x\cos(x^2+5). \end{aligned} $$

Answer:

$$ \boxed{\frac{dy}{dx}=2x\cos(x^2+5).} $$

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Q.2 Differentiate $y=\cos(\sin x)$ with respect to $x$.

Solution:

Given,

$$ y=\cos(\sin x). $$

Using the chain rule,

$$ \frac{d}{dx}(\cos u)=-\sin u\cdot\frac{du}{dx}, $$

where

$$ u=\sin x. $$

Now,

$$ \frac{du}{dx}=\cos x. $$

Hence,

$$ \begin{aligned} \frac{dy}{dx} &=-\sin(\sin x)\cdot\cos x. \end{aligned} $$

Answer:

$$ \boxed{\frac{dy}{dx}=-\cos x\,\sin(\sin x).} $$

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Q.3 Differentiate $y=\sin(ax+b)$ with respect to $x$.

Solution:

Given,

$$ y=\sin(ax+b), $$

where $a$ and $b$ are constants.

Using the chain rule,

$$ \frac{d}{dx}(\sin u)=\cos u\cdot\frac{du}{dx}, $$

where

$$ u=ax+b. $$

Now,

$$ \frac{du}{dx}=a. $$

Therefore,

$$ \begin{aligned} \frac{dy}{dx} &=\cos(ax+b)\cdot a\\ &=a\cos(ax+b). \end{aligned} $$

Answer:

$$ \boxed{\frac{dy}{dx}=a\cos(ax+b).} $$

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Q.4 Differentiate $y=\sec(\tan\sqrt{x})$ with respect to $x$.

Solution:

Given,

$$ y=\sec(\tan\sqrt{x}). $$

Using the chain rule,

$$ \frac{d}{dx}(\sec u)=\sec u\tan u\cdot\frac{du}{dx}, $$

where

$$ u=\tan\sqrt{x}. $$

Now,

$$ \frac{du}{dx} =\sec^2(\sqrt{x})\cdot\frac{d}{dx}(\sqrt{x}). $$

Since

$$ \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}, $$

we get

$$ \begin{aligned} \frac{du}{dx} &=\sec^2(\sqrt{x})\cdot\frac{1}{2\sqrt{x}}. \end{aligned} $$

Therefore,

$$ \begin{aligned} \frac{dy}{dx} &=\sec(\tan\sqrt{x})\tan(\tan\sqrt{x})\cdot \sec^2(\sqrt{x})\cdot\frac{1}{2\sqrt{x}}. \end{aligned} $$

Answer:

$$ \boxed{\frac{dy}{dx} =\frac{\sec(\tan\sqrt{x})\,\tan(\tan\sqrt{x})\,\sec^2(\sqrt{x})}{2\sqrt{x}}.} $$

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Q.5 Differentiate

$$ y=\frac{\sin(ax+b)}{\cos(cx+d)} $$

with respect to $x$, where $a$, $b$, $c$ and $d$ are constants.

Solution:

Given,

$$ y=\frac{\sin(ax+b)}{\cos(cx+d)}. $$

Using the quotient rule,

$$ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}, $$

where

$$ u=\sin(ax+b), \qquad v=\cos(cx+d). $$

Differentiate $u$:

$$ \begin{aligned} \frac{du}{dx} &=\cos(ax+b)\cdot a\\ &=a\cos(ax+b). \end{aligned} $$

Differentiate $v$:

$$ \begin{aligned} \frac{dv}{dx} &=-\sin(cx+d)\cdot c\\ &=-c\sin(cx+d). \end{aligned} $$

Applying the quotient rule,

$$ \begin{aligned} \frac{dy}{dx} &= \frac{\cos(cx+d)\big(a\cos(ax+b)\big) -\sin(ax+b)\big(-c\sin(cx+d)\big)} {\cos^2(cx+d)}\\[2mm] &= \frac{a\cos(ax+b)\cos(cx+d) +c\sin(ax+b)\sin(cx+d)} {\cos^2(cx+d)}. \end{aligned} $$

Answer:

$$ \boxed{ \frac{dy}{dx} = \frac{a\cos(ax+b)\cos(cx+d) +c\sin(ax+b)\sin(cx+d)} {\cos^2(cx+d)} } $$

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Q.6 Differentiate $y=\cos(x^3)\cdot\sin^2(x^5)$ with respect to $x$.

Solution:

Given,

$$ y=\cos(x^3)\cdot\sin^2(x^5). $$

Using the product rule,

$$ \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}, $$

where

$$ u=\cos(x^3), \qquad v=\sin^2(x^5). $$

Differentiate $u$ using the chain rule:

$$ \begin{aligned} \frac{du}{dx} &=-\sin(x^3)\cdot3x^2\\ &=-3x^2\sin(x^3). \end{aligned} $$

Differentiate $v$:

$$ \begin{aligned} \frac{dv}{dx} &=2\sin(x^5)\cdot \frac{d}{dx}\big(\sin(x^5)\big)\\ &=2\sin(x^5)\cos(x^5)\cdot5x^4\\ &=10x^4\sin(x^5)\cos(x^5). \end{aligned} $$

Applying the product rule,

$$ \begin{aligned} \frac{dy}{dx} &= \cos(x^3)\left(10x^4\sin(x^5)\cos(x^5)\right) +\sin^2(x^5)\left(-3x^2\sin(x^3)\right)\\[2mm] &= 10x^4\cos(x^3)\sin(x^5)\cos(x^5) -3x^2\sin(x^3)\sin^2(x^5). \end{aligned} $$

Answer:

$$ \boxed{ \frac{dy}{dx} = 10x^4\cos(x^3)\sin(x^5)\cos(x^5) -3x^2\sin(x^3)\sin^2(x^5) } $$

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Q.7 Differentiate $y=2\sqrt{\cot(x^2)}$ with respect to $x$.

Solution:

Given,

$$ y=2\sqrt{\cot(x^2)}. $$

Using the chain rule,

$$ \frac{d}{dx}(\sqrt{u}) = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}. $$

Let

$$ u=\cot(x^2). $$

Then,

$$ y=2\sqrt{u}. $$

Differentiating,

$$ \begin{aligned} \frac{dy}{dx} &= 2\cdot \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}\\ &= \frac{1}{\sqrt{u}} \cdot \frac{du}{dx}. \end{aligned} $$

Now,

$$ \begin{aligned} \frac{du}{dx} &= -\csc^2(x^2)\cdot2x\\ &= -2x\csc^2(x^2). \end{aligned} $$

Hence,

$$ \begin{aligned} \frac{dy}{dx} &= \frac{-2x\csc^2(x^2)} {\sqrt{\cot(x^2)}}. \end{aligned} $$

Answer:

$$ \boxed{ \frac{dy}{dx} = \frac{-2x\,\csc^2(x^2)} {\sqrt{\cot(x^2)}} } $$

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Q.8 Differentiate $y=\cos(\sqrt{x})$ with respect to $x$.

Solution:

Given,

$$ y=\cos(\sqrt{x}). $$

Using the chain rule,

$$ \frac{d}{dx}(\cos u) = -\sin u\cdot\frac{du}{dx}, $$

where

$$ u=\sqrt{x}. $$

Now,

$$ \frac{du}{dx} = \frac{1}{2\sqrt{x}}. $$

Therefore,

$$ \begin{aligned} \frac{dy}{dx} &= -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}\\ &= -\frac{\sin(\sqrt{x})}{2\sqrt{x}}. \end{aligned} $$

Answer:

$$ \boxed{ \frac{dy}{dx} = -\frac{\sin(\sqrt{x})}{2\sqrt{x}} } $$

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Q.9 Prove that the function given by

$$ f(x)=|x-1|,\qquad x\in\mathbb{R} $$

is not differentiable at $x=1$.

Solution:

The given function is

$$ f(x)=|x-1|. $$

We know that a function is differentiable at $x=a$ if

$$ \lim_{h\to0^-}\frac{f(a+h)-f(a)}{h} = \lim_{h\to0^+}\frac{f(a+h)-f(a)}{h}. $$

We shall examine the left-hand derivative and right-hand derivative at $x=1$.

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Step 1: Find the value of the function at $x=1$.

$$ \begin{aligned} f(1) &=|1-1|\\ &=0. \end{aligned} $$

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Step 2: Find the left-hand derivative.

For $h<0$, we have

$$ 1+h<1. $$

Therefore,

$$ \begin{aligned} f(1+h) &=|(1+h)-1|\\ &=|h|\\ &=-h. \end{aligned} $$

Hence,

$$ \begin{aligned} LHD &=\lim_{h\to0^-}\frac{f(1+h)-f(1)}{h}\\ &=\lim_{h\to0^-}\frac{-h-0}{h}\\ &=\lim_{h\to0^-}(-1)\\ &=-1. \end{aligned} $$

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Step 3: Find the right-hand derivative.

For $h>0$, we have

$$ 1+h>1. $$

Therefore,

$$ \begin{aligned} f(1+h) &=|(1+h)-1|\\ &=|h|\\ &=h. \end{aligned} $$

Hence,

$$ \begin{aligned} RHD &=\lim_{h\to0^+}\frac{f(1+h)-f(1)}{h}\\ &=\lim_{h\to0^+}\frac{h-0}{h}\\ &=\lim_{h\to0^+}1\\ &=1. \end{aligned} $$

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Step 4: Compare the derivatives.

$$ LHD=-1 \ne 1=RHD. $$

Since the left-hand derivative and the right-hand derivative are not equal, the derivative at $x=1$ does not exist.

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Conclusion:

Hence, the function

$$ f(x)=|x-1| $$

is not differentiable at $x=1$.

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Q.10 Prove that the greatest integer function defined by

$$ f(x)=[x],\qquad 0 is not differentiable at $x=1$ and $x=2$.

Solution:

The greatest integer function $[x]$ has jump discontinuities at every integer.

Since differentiability implies continuity, a function that is discontinuous at a point cannot be differentiable there.

We verify this using the definition of derivative.

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(i) At $x=1$

Since

$$ f(1)=[1]=1, $$

the left-hand derivative is

$$ \begin{aligned} LHD &=\lim_{h\to0^-}\frac{[1+h]-1}{h}. \end{aligned} $$

For $h<0$, we have

$$ 0<1+h<1, $$

so

$$ [1+h]=0. $$

Hence,

$$ \begin{aligned} LHD &=\lim_{h\to0^-}\frac{0-1}{h}\\ &=\lim_{h\to0^-}\frac{-1}{h}\\ &=+\infty. \end{aligned} $$

The right-hand derivative is

$$ \begin{aligned} RHD &=\lim_{h\to0^+}\frac{[1+h]-1}{h}. \end{aligned} $$

For $h>0$,

$$ 1<1+h<2, $$

so

$$ [1+h]=1. $$

Therefore,

$$ \begin{aligned} RHD &=\lim_{h\to0^+}\frac{1-1}{h}\\ &=0. \end{aligned} $$

Since

$$ LHD\ne RHD, $$

the function is not differentiable at $x=1$.

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(ii) At $x=2$

Since

$$ f(2)=[2]=2, $$

the left-hand derivative is

$$ \begin{aligned} LHD &=\lim_{h\to0^-}\frac{[2+h]-2}{h}. \end{aligned} $$

For $h<0$,

$$ 1<2+h<2, $$

therefore,

$$ [2+h]=1. $$

Hence,

$$ \begin{aligned} LHD &=\lim_{h\to0^-}\frac{1-2}{h}\\ &=\lim_{h\to0^-}\frac{-1}{h}\\ &=+\infty. \end{aligned} $$

The right-hand derivative is

$$ \begin{aligned} RHD &=\lim_{h\to0^+}\frac{[2+h]-2}{h}. \end{aligned} $$

For $h>0$,

$$ 2<2+h<3, $$

so

$$ [2+h]=2. $$

Therefore,

$$ \begin{aligned} RHD &=\lim_{h\to0^+}\frac{2-2}{h}\\ &=0. \end{aligned} $$

Since

$$ LHD\ne RHD, $$

the function is not differentiable at $x=2$.

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Conclusion:

Hence, the greatest integer function

$$ f(x)=[x] $$

is not differentiable at $x=1$ and $x=2$.

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