Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$\sqrt{\frac{1+\sin x}{1-\sin x}}$

Solution:

Let $y=\sqrt{\frac{1+\sin x}{1-\sin x}}$

On differentiating y with respect to $x$, we get

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\left(\frac{1+\sin \mathrm{x}}{1-\sin \mathrm{x}}\right)^{\frac{1}{2}}\right]$

We know $\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\frac{1+\sin \mathrm{x}}{1-\sin \mathrm{x}}\right)^{\frac{1}{2}-1} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1+\sin x}{1-\sin x}\right)$ [using chain rule]

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left(\frac{1+\sin x}{1-\sin x}\right)^{-\frac{1}{2}} \frac{d}{d x}\left(\frac{1+\sin x}{1-\sin x}\right)$

Recall that $\left(\frac{\mathrm{u}}{\mathrm{v}}\right)^{\prime}=\frac{\mathrm{vu}^{\prime}-\mathrm{uv}^{\prime}}{\mathrm{v}^{2}}$ (quotient rule)

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\frac{1+\sin \mathrm{x}}{1-\sin \mathrm{x}}\right)^{-\frac{1}{2}}\left[\frac{(1-\sin \mathrm{x}) \frac{\mathrm{d}}{\mathrm{dx}}(1+\sin \mathrm{x})-(1+\sin \mathrm{x}) \frac{\mathrm{d}}{\mathrm{dx}}(1-\sin \mathrm{x})}{(1-\sin \mathrm{x})^{2}}\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}$

$=\frac{1}{2}\left(\frac{1+\sin x}{1-\sin x}\right)^{-\frac{1}{2}}\left[\frac{(1-\sin x)\left(\frac{d}{d x}(1)+\frac{d}{d x}(\sin x)\right)-(1+\sin x)\left(\frac{d}{d x}(1)-\frac{d}{d x}(\sin x)\right)}{(1-\sin x)^{2}}\right]$

We know $\frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})=\cos \mathrm{x}$ and derivative of a constant is 0 .

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left(\frac{1+\sin x}{1-\sin x}\right)^{-\frac{1}{2}}\left[\frac{(1-\sin x)(0+\cos x)-(1+\sin x)(0-\cos x)}{(1-\sin x)^{2}}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left(\frac{1+\sin x}{1-\sin x}\right)^{-\frac{1}{2}}\left[\frac{(1-\sin x) \cos x+(1+\sin x) \cos x}{(1-\sin x)^{2}}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left(\frac{1+\sin x}{1-\sin x}\right)^{-\frac{1}{2}}\left[\frac{(1-\sin x) \cos x+(1+\sin x) \cos x}{(1-\sin x)^{2}}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left(\frac{1+\sin x}{1-\sin x}\right)^{-\frac{1}{2}}\left[\frac{(1-\sin x+1+\sin x) \cos x}{(1-\sin x)^{2}}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left(\frac{1+\sin x}{1-\sin x}\right)^{-\frac{1}{2}}\left[\frac{2 \cos x}{(1-\sin x)^{2}}\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\left(\frac{1+\sin \mathrm{x}}{1-\sin \mathrm{x}}\right)^{-\frac{1}{2}}\left[\frac{\cos \mathrm{x}}{(1-\sin \mathrm{x})^{2}}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{(1+\sin x)^{-\frac{1}{2}}}{(1-\sin x)^{-\frac{1}{2}}}\left[\frac{\cos x}{(1-\sin x)^{2}}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{(1+\sin x)^{-\frac{1}{2}} \cos x}{(1-\sin x)^{\frac{1}{2}+2}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{(1+\sin \mathrm{x})^{-\frac{1}{2}} \cos \mathrm{x}}{(1-\sin \mathrm{x})^{\frac{3}{2}}}$

$\Rightarrow \frac{d y}{d x}=\frac{\cos x}{(1-\sin x)^{1+\frac{1}{2}}(1+\sin x)^{\frac{1}{2}}}$

$\Rightarrow \frac{d y}{d x}=\frac{\cos x}{(1-\sin x)(1-\sin x)^{\frac{1}{2}}(1+\sin x)^{\frac{1}{2}}}$

$\Rightarrow \frac{d y}{d x}=\frac{\cos x}{(1-\sin x) \sqrt{(1-\sin x)(1+\sin x)}}$

$\Rightarrow \frac{d y}{d x}=\frac{\cos x}{(1-\sin x) \sqrt{1-\sin ^{2} x}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\cos \mathrm{x}}{(1-\sin \mathrm{x}) \sqrt{\cos ^{2} \mathrm{x}}}\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

$\Rightarrow \frac{d y}{d x}=\frac{\cos x}{(1-\sin x) \cos x}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{1-\sin x}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{1-\sin x} \times \frac{1+\sin x}{1+\sin x}$

$\Rightarrow \frac{d y}{d x}=\frac{1+\sin x}{1-\sin ^{2} x}$

$\Rightarrow \frac{d y}{d x}=\frac{1+\sin x}{\cos ^{2} x}\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\cos ^{2} x}+\frac{\sin x}{\cos ^{2} x}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\left(\frac{1}{\cos \mathrm{x}}\right)^{2}+\left(\frac{1}{\cos \mathrm{x}}\right)\left(\frac{\sin \mathrm{x}}{\cos \mathrm{x}}\right)$

$\Rightarrow \frac{d y}{d x}=\sec ^{2} x+\sec x \tan x$

$\therefore \frac{d y}{d x}=\sec x(\sec x+\tan x)$

Thus, $\frac{d}{d x}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)=\sec x(\sec x+\tan x)$

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