Exercise 5.4

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Q.1 Differentiate

$$ y=\frac{e^x}{\sin x} $$

with respect to $x$.

Solution:

Given,

$$ y=\frac{e^x}{\sin x}. $$

Using the quotient rule,

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

where

$$ u=e^x, \qquad v=\sin x. $$

Now,

$$ \frac{du}{dx}=e^x, \qquad \frac{dv}{dx}=\cos x. $$

Substituting in the quotient rule,

$$ \begin{aligned} \frac{dy}{dx} &= \frac{\sin x\,(e^x)-e^x(\cos x)} {\sin^2x}\\[2mm] &= \frac{e^x(\sin x-\cos x)} {\sin^2x}. \end{aligned} $$

Answer:

$$ \boxed{ \frac{dy}{dx} = \frac{e^x(\sin x-\cos x)} {\sin^2x} } $$

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Q.2 Differentiate

$$ y=e^{\sin^{-1}x} $$

with respect to $x$.

Solution:

Given,

$$ y=e^{\sin^{-1}x}. $$

Using the chain rule,

$$ \frac{d}{dx}(e^u) = e^u\frac{du}{dx}, $$

where

$$ u=\sin^{-1}x. $$

Now,

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

Therefore,

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

Answer:

$$ \boxed{ \frac{dy}{dx} = \frac{e^{\sin^{-1}x}} {\sqrt{1-x^2}} } $$

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Q.3 Differentiate

$$ y=e^{x^3} $$

with respect to $x$.

Solution:

Given,

$$ y=e^{x^3}. $$

Using the chain rule,

$$ \frac{d}{dx}(e^u) = e^u\frac{du}{dx}, $$

where

$$ u=x^3. $$

Now,

$$ \frac{du}{dx}=3x^2. $$

Therefore,

$$ \begin{aligned} \frac{dy}{dx} &= e^{x^3}\cdot3x^2\\ &= 3x^2e^{x^3}. \end{aligned} $$

Answer:

$$ \boxed{ \frac{dy}{dx} = 3x^2e^{x^3} } $$

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Q.4 Differentiate

$$ y=\sin\left(\tan^{-1}e^{-x}\right) $$

with respect to $x$.

Solution:

Given,

$$ y=\sin\left(\tan^{-1}e^{-x}\right). $$

Using the chain rule,

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

where

$$ u=\tan^{-1}e^{-x}. $$

Now,

$$ \frac{d}{dx}(\tan^{-1}v) = \frac{1}{1+v^2}\cdot\frac{dv}{dx}, $$

where

$$ v=e^{-x}. $$

Since

$$ \frac{dv}{dx}=-e^{-x}, $$

we obtain

$$ \begin{aligned} \frac{du}{dx} &= \frac{-e^{-x}} {1+e^{-2x}}. \end{aligned} $$

Therefore,

$$ \begin{aligned} \frac{dy}{dx} &= \cos\left(\tan^{-1}e^{-x}\right) \left( \frac{-e^{-x}} {1+e^{-2x}} \right). \end{aligned} $$

Using

$$ \cos(\tan^{-1}t)=\frac{1}{\sqrt{1+t^2}}, $$

with $t=e^{-x}$,

$$ \cos\left(\tan^{-1}e^{-x}\right) = \frac{1}{\sqrt{1+e^{-2x}}}. $$

Hence,

$$ \boxed{ \frac{dy}{dx} = -\frac{e^{-x}} {\left(1+e^{-2x}\right)^{3/2}} } $$

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Q.5 Differentiate

$$ y=\log(\cos e^x) $$

with respect to $x$.

Solution:

Given,

$$ y=\log(\cos e^x). $$

Using the chain rule,

$$ \frac{d}{dx}(\log u) = \frac{1}{u}\cdot\frac{du}{dx}, $$

where

$$ u=\cos e^x. $$

Now,

$$ \begin{aligned} \frac{du}{dx} &= -\sin(e^x)\cdot e^x. \end{aligned} $$

Therefore,

$$ \begin{aligned} \frac{dy}{dx} &= \frac{-e^x\sin(e^x)} {\cos(e^x)}\\[2mm] &= -e^x\tan(e^x). \end{aligned} $$

Answer:

$$ \boxed{ \frac{dy}{dx} = -e^x\tan(e^x) } $$

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Q.6 Differentiate

$$ y=e^x+e^{x^2}+e^{x^3}+\cdots+e^{x^5} $$

with respect to $x$.

Solution:

Given,

$$ y=e^x+e^{x^2}+e^{x^3}+e^{x^4}+e^{x^5}. $$

Differentiate each term separately using the chain rule.

We know that

$$ \frac{d}{dx}(e^{x^n}) = ne^{x^n}x^{n-1}. $$

Therefore,

$$ \begin{aligned} \frac{dy}{dx} &= e^x + 2xe^{x^2} + 3x^2e^{x^3} + 4x^3e^{x^4} + 5x^4e^{x^5}. \end{aligned} $$

Answer:

$$ \boxed{ \frac{dy}{dx} = e^x + 2xe^{x^2} + 3x^2e^{x^3} + 4x^3e^{x^4} + 5x^4e^{x^5} } $$

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Q.7 Differentiate

$$ y=\sqrt{e^{\sqrt{x}}}, \qquad x>0 $$

with respect to $x$.

Solution:

Given,

$$ y=\sqrt{e^{\sqrt{x}}}. $$

We can write

$$ y=\left(e^{\sqrt{x}}\right)^{\frac12}. $$

Using the chain rule,

$$ \frac{d}{dx}\left(u^{\frac12}\right) = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}, $$

where

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

Now,

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

Since

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

we get

$$ \begin{aligned} \frac{du}{dx} &= \frac{e^{\sqrt{x}}}{2\sqrt{x}}. \end{aligned} $$

Therefore,

$$ \begin{aligned} \frac{dy}{dx} &= \frac{1}{2\sqrt{e^{\sqrt{x}}}} \cdot \frac{e^{\sqrt{x}}}{2\sqrt{x}}\\[2mm] &= \frac{\sqrt{e^{\sqrt{x}}}} {4\sqrt{x}}. \end{aligned} $$

Answer:

$$ \boxed{ \frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}} {4\sqrt{x}} } $$

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Q.8 Differentiate

$$ y=\log(\log x), \qquad x>1 $$

with respect to $x$.

Solution:

Given,

$$ y=\log(\log x). $$

Using the chain rule,

$$ \frac{d}{dx}(\log u) = \frac{1}{u}\cdot\frac{du}{dx}, $$

where

$$ u=\log x. $$

Now,

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

Therefore,

$$ \begin{aligned} \frac{dy}{dx} &= \frac{1}{\log x} \cdot \frac{1}{x}\\[2mm] &= \frac{1}{x\log x}. \end{aligned} $$

Answer:

$$ \boxed{ \frac{dy}{dx} = \frac{1}{x\log x} } $$

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Q.9 Differentiate

$$ y=\frac{\cos x}{\log x}, \qquad x>0 $$

with respect to $x$.

Solution:

Given,

$$ y=\frac{\cos x}{\log x}. $$

Using the quotient rule,

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

where

$$ u=\cos x, \qquad v=\log x. $$

Now,

$$ \frac{du}{dx}=-\sin x, \qquad \frac{dv}{dx}=\frac{1}{x}. $$

Applying the quotient rule,

$$ \begin{aligned} \frac{dy}{dx} &= \frac{\log x(-\sin x)-\cos x\left(\frac1x\right)} {(\log x)^2}\\[2mm] &= \frac{-x\log x\sin x-\cos x} {x(\log x)^2}. \end{aligned} $$

Answer:

$$ \boxed{ \frac{dy}{dx} = -\frac{x\log x\,\sin x+\cos x} {x(\log x)^2} } $$

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Q.10 Differentiate

$$ y=\cos(\log x+e^x), \qquad x>0 $$

with respect to $x$.

Solution:

Given,

$$ y=\cos(\log x+e^x). $$

Using the chain rule,

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

where

$$ u=\log x+e^x. $$

Now,

$$ \begin{aligned} \frac{du}{dx} &= \frac{d}{dx}(\log x) + \frac{d}{dx}(e^x)\\ &= \frac1x+e^x. \end{aligned} $$

Therefore,

$$ \begin{aligned} \frac{dy}{dx} &= -\sin(\log x+e^x) \left(\frac1x+e^x\right). \end{aligned} $$

Answer:

$$ \boxed{ \frac{dy}{dx} = -\left(\frac1x+e^x\right) \sin(\log x+e^x) } $$

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