Question

y=log(sin(tanx))
$y = log(sin(tanx)) \: find \frac{dy}{dx}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:y = log\bigg[sin(tanx)\bigg]$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx} log\bigg[sin(tanx)\bigg]$

We know,

$\red{\boxed{ \sf \:\dfrac{d}{dx}logx = \frac{1}{x}}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{sin(tanx)} \: \dfrac{d}{dx} \: sin(tanx)$

We know,

$\red{\boxed{ \sf \:\dfrac{d}{dx}sinx = cosx}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{sin(tanx)} \: \times \: cos(tanx) \: \dfrac{d}{dx} \: tanx$

We know,

$\red{\boxed{ \sf \:\dfrac{d}{dx}tanx = {sec}^{2}x}}$

So, using we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = cot(tanx) \times {sec}^{2}x$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \sf \: \: \: \: \: \:\dfrac{dy}{dx} = cot(tanx) \times {sec}^{2}x \: \: \: \: \: \: }}$

Additional Information :-

$\red{\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}}$