Question

If y=logtanx prove that dy/dx = 2 cosec2x​

Answer 1

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

Given function is

$\rm :\longmapsto\:y \: = \: log \: tanx$

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

$\rm :\longmapsto\:\dfrac{d}{dx} y \: = \dfrac{d}{dx} \: log \: tanx$

We know

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

So, using this, we get

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{tanx} \times {sec}^{2}x$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{1}{\dfrac{sinx}{cosx}} \times \dfrac{1}{ {cos}^{2} x}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{sinx \: cosx}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{2}{2 \: sinx \: cosx}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{2}{sin2x}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 2 \: cosec2x$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{dy}{dx} = 2 \: cosec2x \: }}$

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Additional Information :-

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