Question

Differentiate the function
${e}^{6x } log( \tan(x) )$ with respective to x.

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Answer 1

Answer:

Derivative of

$\rm \: {e}^{ 6x} log \tan x$

Here we go,

Solution :-

Let:-

$\rm y = \: {e}^{ 6x} log \tan x$

Differentiating both sides with respect to x,we get

$\rm \dfrac{dy}{dx} = \dfrac{d}{dx}( {e}^{6x} log \tan x)$

$\rm\dfrac{dy}{dx} = {e}^{6x} \dfrac{d}{dx} ( log \tan x) + log \tan x. \dfrac{d}{dx} ( {e}^{6x} )$

$\rm\dfrac{dy}{dx} = {e}^{6x} . \dfrac{1}{ \tan x}. \dfrac{d}{dx} ( \tan x) + log \tan x. {e}^{6x} . \frac{d}{dx} (6x)$

$\rm\dfrac{dy}{dx} = {e}^{6x}. \dfrac{1}{ \tan x} . { \sec }^{2} x + log \tan x. {e}^{x} .6$

$\rm\dfrac{dy}{dx} = {e}^{6x}. \dfrac{ \cancel{\cos x}}{ \sin x } . \dfrac{1}{ { \cos }^{ \cancel2} x} +6{e}^{x} log \tan x$

$\rm\dfrac{dy}{dx} = {e}^{6x} \csc x. \sec x+6{e}^{x} log \tan x$

$\rm\dfrac{dy}{dx} = 6{e}^{6x}( log\tan x + \dfrac{1}{6} \csc x. \sec x)$

Important Derivative Formulae

$\boxed{\begin{array}{c|c} \rm \: \underline{function}& \rm \underline{Derivative} \\ \\ \rm \dfrac{d}{dx} ( {e}^{x} ) \: \: \: \: \: \: \: \: \: \ & \rm {e}^{x} \\ \\ \rm \: \dfrac{d}{dx}( log x ) \: \: \: \: \: \: \: \: & \rm \dfrac{1}{x}\\ \\ \rm \dfrac{d}{dx}( \sin x )\: \: \: \: \: \: & \rm \cos x \\ \\ \rm \dfrac{d}{dx}( \cos x ) \: \: \: & \rm - \sin x \\ \\ \rm \dfrac{d}{dx}( \tan x ) & \rm \: { \sec}^{2}x \\ \\ \rm \dfrac{d}{dx}( \cot x ) & \rm- { \csc }^{2}x \\ \\ \rm \dfrac{d}{dx}( \sec x ) & \rm \sec x. \tan x \\ \\\rm \dfrac{d}{dx}( \csc x ) & \rm \: - \csc x. \cot x\\ \\ \rm \dfrac{d}{dx}(x) \: \: \: \: \: \: \: & 1 \end{array}}$

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