Question

iii) Differentiate log[sec(2x)] w.rito x .​

Answer 1

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

Given function is

$\red{\rm :\longmapsto\:log[sec(2x)] \: }$

Let assume that

${\rm :\longmapsto\:y = log[sec(2x)] \: }$

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

${\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx} log[sec(2x)] \: }$

We know

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

So, using this, we get

${\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{1}{sec2x} \: \dfrac{d}{dx} \: sec(2x)\: }$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}secx \: = \: secx \: tanx \: }}$

So, using this, we get

${\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{1}{sec2x} \: sec2x \: tan2x \: \dfrac{d}{dx}(2x)\: }$

${\rm :\longmapsto\:\dfrac{dy}{dx} =2 \: tan2x \: \dfrac{d}{dx}(x)\: }$

${\rm :\longmapsto\:\dfrac{dy}{dx} =2 \: tan2x \: \times 1\: }$

${\rm :\longmapsto\:\dfrac{dy}{dx} =2 \: tan2x \: \: }$

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More to know :-

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