Question

differentiate log sinx/2​

Answer 1

Question :

Differentiate $\tt log(sin\frac{x}{2})$

Solution :

Let y = $\tt log(sin\frac{x}{2})$

Differentiating w.r.t x

$\tt \implies \frac{dy}{dx} = \frac{d(log(sin\frac{x}{2})}{dx}$

$\tt \implies \frac{dy}{dx} = \frac{1}{sin\frac{x}{2}} \frac{d(sin\frac{x}{2})}{dx}$

$\tt \implies \frac{dy}{dx} = \frac{cos\frac{x}{2}}{sin\frac{x}{2}} \frac{d(\frac{x}{2})}{dx}$

$\tt \implies \frac{dy}{dx} = cot\frac{x}{2} \frac{1}{2}$

$\tt \implies \frac{dy}{dx} = \frac{1}{2}cot\frac{x}{2}$

Formulas in Differentiation :

● $\tt \frac{d({x}^{n})}{dx} = n{x}^{n-1}$

● $\tt \frac{d(logx)}{dx} = \frac{1}{x}$

●$\tt \frac{d({e}^{x})}{dx} = {e}^{x}$

● $\tt \frac{d(sinx)}{dx} = cosx$

●$\tt \frac{d(cosx)}{dx} = -sinx$

● $\tt \frac{d(tanx)}{dx} = {sec}^{2}x$

● $\tt \frac{d(cotx)}{dx} = {cosec}^{2}x$

● $\tt \frac{d(secx)}{dx} = secx.tanx$

● $\tt \frac{d(cosecx)}{dx} = - cosecx.cotx$

Answer 2

Answer:-

Let y=

$(log( sinx)) ^{2}$

$= > \frac{dy}{dx} = 2 log( sinx). \frac{d}{dx} (log(sinx)$

$= > 2log(sinx) \times \frac{1}{sin} \times \frac{d}{dx} (sinx)$

$= > 2log(sinx) \times \frac{1}{sinx} \times cosx$

$= > \frac{2cos \: log(sinx)}{sinx}$