$\frac{\sin x}{(1+\cos x)}$



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

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) $\frac{d \cos x}{d x}=-\sin x$

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

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

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

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

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)

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

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

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

$=\frac{1}{(1+\cos x)}$

Ans) $=\frac{1}{1+\cos x}$


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