EXERCISE 5.4 ( Differentiation )

Differentiating the following w.r.t.(Class 12 ncert solution math exercise 5.4)

Question 1: $\frac{e^x}{\sin x}$ .

Solution:Let $y=\frac{e^x}{\sin x}$

By using the quotient rule, we get

$\frac{dy}{dx} =\frac{\sin x \frac{d}{d x}\left(e^x\right)-e^x \frac{d}{d x}(\sin x)}{\sin ^2 x}$

$=\frac{\sin x \cdot\left(e^x\right)-e^x \cdot(\cos x)}{\sin ^2 x}$

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

Question 2: $e^{\sin ^{-1} x}$

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

By using the quotient rule, we get

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

$=e^{\sin ^{-1} x} \cdot \frac{d}{d x}\left(\sin ^{-1} x\right)$

$=e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}$

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

$=\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^2}}, x \in(-1,1)$

Question 3: $e^{x^3}$

Solution:Let $y=e^{x^3}$

By using the quotient rule, we get

$\frac{d y}{d x} =\frac{d}{d x}\left(e^{x^3}\right)$

$=e^{x^3} \cdot \frac{d}{d x}\left(x^3\right)$

$=e^{x^3} \cdot 3 x^2$

$=3 x^2 e^{x^3}$

Question 4: $\sin \left(\tan ^{-1} e^{-x}\right)$

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

By using the chain rule, we get

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

$=\cos \left(\tan ^{-1} e^{-x}\right) \cdot \frac{d}{d x}\left(\tan ^{-1} e^{-x}\right)$

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

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

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

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

Question 5: $\log \left(\cos e^x\right)$

Solution:Let $y=\log \left(\cos e^x\right)$

By using the chain rule, we get

$\frac{d y}{d x} =\frac{d}{d x}\left[\log \left(\cos e^x\right)\right]$

$=\frac{1}{\cos e^x} \cdot \frac{d}{d x}\left(\cos e^x\right)$

$=\frac{1}{\cos e^x} \cdot\left(-\sin e^x\right) \cdot \frac{d}{d x}\left(e^x\right)$

$=\frac{-\sin e^x}{\cos e^x} \cdot e^x$

$=-e^x \tan e^x, e^x \neq(2 n+1) \frac{\pi}{2}, n \in \mathbf{N}$

Question 6: $e^x+e^{x^2}+\ldots+e^{x^5}$

Solution: $\frac{d}{d x}\left(e^x+e^{x^2}+\ldots+e^{x^5}\right)$

Differentiating wrt $x$ , we get

$\frac{d}{d x}\left(e^x+e^{x^2}+\ldots+e^{x^5}\right) =\frac{d}{d x}\left(e^x\right)+\frac{d}{dx}\left(e^{x^2}\right)+\frac{d}{d x}\left(e^{x^3}\right)+\frac{d}{d x}\left(e^{x^4}\right)+\frac{d}{d x}\left(e^{x^5}\right)$

$=e^x+\left[e^{x^2} \times \frac{d}{d x}\left(x^2\right)\right]+\left[e^{x^3} \times \frac{d}{d x}\left(x^3\right)\right]+\left[e^{x^4} \times \frac{d}{d x}\left(x^4\right)\right]+\left[e^{x^5} \times \frac{d}{d x}\left(x^5\right)\right]$