# Ncert Solutions For Class 12 Maths Ex 5.4

## Ncert Solutions For Class 12 Maths Chapter 5 Ex 5.4

Q1: Differentiate the following w.r.t. x

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

Sol:

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

Using quotient rule, we have

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\sin x.\frac{\mathrm{d} }{\mathrm{d} x}(e^{x})-e^{x}.\frac{\mathrm{d} }{\mathrm{d} x}(\sin x)}{\sin^{2} x}$$ $$=\frac{\sin x.(e^{x})-e^{x}.(\cos x)}{\sin^{2} x}$$ $$=\frac{e^{x}(\sin x-\cos x)}{\sin^{2} x},x\neq n\pi ,n\in \mathbb{Z}$$

Q2: Differentiate the following w.r.t. x

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

Sol:

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

Using chain rule, we have

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(e^{\sin ^{-1}x})$$ $$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=e^{\sin^{-1}x}.\frac{\mathrm{d} }{\mathrm{d} x}(\sin^{-1}x)$$ $$=e^{\sin^{-1}x}.\frac{1}{\sqrt{1-x^{2}}}$$ $$=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}}$$ $$therefore,\; \frac{\mathrm{d} y}{\mathrm{d}x}=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}},x\in (-1,1)$$

Q3: Differentiate the following w.r.t. x

$$e^{x^{3}}$$

Sol:

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

Using chain rule, we have

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(e^{x^{3}})=e^{x^{3}}.3x^{2}=3x^{2}.e^{x^{3}}$$

Q4: Differentiate the following w.r.t. x

$$\sin (\tan^{-1} e^{-x})$$

Sol:

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

Using chain rule, we have

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}[\sin (\tan^{-1} e^{-x})]$$ $$=\cos (\tan^{-1} e^{-x}).\frac{\mathrm{d} }{\mathrm{d} x}(\tan^{-1} e^{-x})$$ $$=\cos (\tan^{-1} e^{-x}).\frac{1}{1+(e^{-x})^{2}}(\tan^{-1} e^{-x}).\frac{\mathrm{d} }{\mathrm{d} x}(e^-{x})$$ $$=\frac{\cos (\tan^{-1} e^{-x})}{1+e^{-2x}}.e^{-x}.\frac{\mathrm{d} }{\mathrm{d} x}(-x)$$ $$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=e^{\sin^{-1}x}.\frac{\mathrm{d} }{\mathrm{d} x}(\sin^{-1}x)$$ $$=e^{\sin^{-1}x}.\frac{1}{\sqrt{1-x^{2}}}$$ $$=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}}$$ $$therefore,\; \frac{\mathrm{d} y}{\mathrm{d}x}=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}},x\in (-1,1)$$