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Differentiate

Question:

Differentiate

$\left(\frac{1+\sin x}{1-\sin x}\right)$

Solution:

To find: Differentiation of $\left(\frac{1+\sin x}{1-\sin x}\right)$

Formula used: (i) $\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

(ii) $\frac{d \sin x}{d x}=\cos x$

Let us take $u=(1+\sin x)$ and $v=(1-\sin x)$

$u^{\prime}=\frac{d u}{d x}=\frac{d(1+\sin x)}{d x}=\cos x$

$v^{\prime}=\frac{d v}{d x}=\frac{d(1-\sin x)}{d x}=-\cos x$

Putting the above obtained values in the formula:-

$\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

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

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

$=\frac{2 \cos x}{(1-\sin x)^{2}}$

Ans $)=\frac{2 \cos x}{(1-\sin x)^{2}}$

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