Question

Y = e^x log tan 2x , find dy/dx

Answer 1

Answer:

${e}^{x} \: ( \: 4 \: cosec4x + log \: tan2x \: )$

Step-by-step explanation:

To find ---->

$derivative \: of \: \\ y \: = {e}^{x} \: \: log \: tan2x$

Concept used ---->

1)

$\dfrac{d}{dx} \: (u \: v ) =u \: \dfrac{dv}{dx} \: + v \: \dfrac{du}{dx}$

2)

$\dfrac{d}{dx} \: ( {e}^{x} ) = {e}^{x} \\\\ \dfrac{d}{dx} (logx) = \dfrac{1}{x} \\ \\ \dfrac{d}{dx} (tanx) = {sec}^{2} x$

3)

$sin2x = 2 \: sinx \: cosx$

Solution---->

$y \: = {e}^{x} \: log \: tan2x$

$differentiating \: with \: respect \: to \: x$

$\dfrac{dy}{dx} \: = {e}^{x} \dfrac{d}{dx} (logtan2x) + logtan2x \: \dfrac{d}{dx} ( {e}^{x} )$

$= {e}^{x} \: \dfrac{1}{tan2x} \: \dfrac{d}{dx} (tan2x) \: + logtan2x \: {e}^{x}$

$= \dfrac{ {e}^{x} \: {sec}^{2}2x \: \dfrac{d}{dx} (2x) }{tan2x} \: + {e}^{x} \: logtan2x$

$= {e}^{x} ( \dfrac{2 {sec}^{2}2x }{tan2x} \: + logtan2x \: )$

$= {e}^{x} ( \dfrac{ \dfrac{2}{ {cos}^{2}2x } }{ \dfrac{sin2x}{cos2x} } \: + logtan2x \: )$

$= {e}^{x} ( \dfrac{2}{sin2x \: cos2x} \: + logtan2x \: )$

$= {e}^{x} \: ( \: \dfrac{4}{2sin2x \: cos2x} + \: logtan2x \: )$

$= {e}^{x} \: ( \: \dfrac{4}{sin4x \: } + \: logtan2x \: )$

$= {e}^{x} \: ( \: 4 \: cosec4x \: + logtan2x \: )$

Answer 2

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