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Answer 1

Given Question :-

$\sf \: \dfrac{d}{dx}log(tanx)$

$\sf \: \: \: \: \: (a) \: 2 \: sec2x$

$\sf \: \: \: \: \: (b) \: 2 \: cosec2x$

$\sf \: \: \: \: \: (c) \: \: sec2x$

$\sf \: \: \: \: \: (d) \: \: cosec2x$

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

Given function is

$\rm :\longmapsto\:\dfrac{d}{dx}log(tanx)$

We know,

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

So, using this, we get

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

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

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

$\rm \:  =  \: \dfrac{1}{sinx \: cosx}$

$\rm \:  =  \: \dfrac{2}{2 \: sinx \: cosx}$

$\rm \:  =  \: \dfrac{2}{sin \: 2x}$

$\rm \:  =  \: 2cosec2x$

Hence,

$\red{\rm \implies\:\rm \:\boxed{\tt{ \dfrac{d}{dx}log(tanx)  =  \: 2cosec2x \: }}}$

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Learn to More

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

Answer 2

Answer:

$</p><p> \mathtt \red{Given \: Question  \: is</p><p>\frac{d}{dx} ( log( \tan(x) ) }$

$\mathtt \red{apply \: chain \: rule : \frac{1}{ \tan(x) } \frac{d}{dx} \tan(x) }$

$\mathtt \red{\implies \: \frac{1}{ \tan(x) } \sec ^{2} (x) }$