Question

find the derivative of cosx^tanx=

Answer 1

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ANSWER:

$The \: \: derivative \: \: of \\ (cosx)^{tanx} = (\log ( \cos x) - { \sin}^{2} x)(cosx)^{ \tan x - 2}$

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GIVEN:

${(cosx)}^{tanx}$

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TO FIND:

$The \: \: derivative \: \: of \: \: (cosx)^{tanx}$

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EXPLANATION:

$Let \: \: y = {(cosx)}^{tanx}$

Take log on both sides

$\log y = \log {( \cos x)}^{ \tan x}$

$\boxed{ \large{ \bold{ \log x^{y} = y \log x }}}$

$\log y = \tan x \log ( \cos x)$

Differentiate w.r.t x

$\boxed{ \large {\bold{\dfrac{d}{dx} ( \log x) = \dfrac{1}{x} }}}$

$\boxed{ \large {\bold{\dfrac{d}{dx} (uv) = uv' + vu' }}}$

$\boxed{ \large {\bold{\dfrac{d}{dx} ( \tan x) = \sec^{2}x}}}$

$\left( \dfrac{1}{y} \right)\dfrac{dy}{dx} = \tan x\left( \dfrac{1}{ \cos x}\right) \dfrac{d}{dx} ( \cos x) + \log ( \cos x)( { \sec}^{2} x)$

$\boxed{ \large {\bold{\dfrac{d}{dx} ( \cos x) = - \sin x}}}$

$\left( \dfrac{1}{y} \right)\dfrac{dy}{dx} = \tan x\left( \dfrac{1}{ \cos x}\right) ( - \sin x) + \log ( \cos x)( { \sec}^{2} x)$

$\boxed{ \large {\bold{\dfrac{ - \sin x}{ \cos x} = - \tan x}}}$

$\left( \dfrac{1}{y} \right)\dfrac{dy}{dx} = \tan x( - \tan x) + \log ( \cos x)( { \sec}^{2} x)$

$\left( \dfrac{1}{y} \right)\dfrac{dy}{dx} = \log ( \cos x)( { \sec}^{2} x) - { \tan}^{2} x$

$\boxed{ \large {\bold{\sec^{2} x= \frac{1}{ { \cos}^{2}x }} }}$

$\boxed{ \large {\bold{\tan^{2} x= \frac{ { \sin}^{2}x }{ { \cos}^{2}x }} }}$

$\left( \dfrac{1}{y} \right)\dfrac{dy}{dx} = \log ( \cos x)\dfrac{1}{ \cos^{2}x } - \dfrac{{ \sin}^{2} x}{ \cos^{2}x }$

$\dfrac{dy}{dx} = y \dfrac{(\log ( \cos x) - { \sin}^{2} x)}{ \cos^{2} x}$

$\boxed{ \large {\bold{{ \frac{{y}^{x} }{ {y}^{2} } } = {y}^{x - 2} } }}$

$\dfrac{y}{ { \cos}^{2} x} = \dfrac{{(cosx)}^{tanx} }{ { \cos}^{2} x} \\ \\ \\ \dfrac{y}{ { \cos}^{2} x} = (cosx)^{tanx - 2}$

$\dfrac{dy}{dx} = (\log ( \cos x) - { \sin}^{2} x)(cosx)^{ \tan x - 2}$

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