Question

find dy/dx if y=x^tanx​

Answer 1

$\rm \longrightarrow y = {x}^{tan \: x}$

Taking log both side :-

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

$\rm \longrightarrow \: log \: y =tan \: x \: \: \: log \: {x}$

$\rm \longrightarrow \: \dfrac{d(log \: y)}{dx} =\dfrac{d(tan \: x \: \: \: log \: {x})}{dx}$

$\rm \longrightarrow \: \dfrac{1}{y} \times \dfrac{dy}{dx} =log \: {x} \: \dfrac{d(tan \: x )}{dx} + tan \: x \: \dfrac{d(log \: x )}{dx}$

$\rm \longrightarrow \: \dfrac{1}{y} \times \dfrac{dy}{dx} =(log \: {x} \times {sec}^{2} x) +( tan \: x \times \dfrac{1}{x} )$

$\rm \longrightarrow \: \dfrac{1}{y} \times \dfrac{dy}{dx} = {sec}^{2} x \: \: log \: {x} + \dfrac{tan \: x}{x}$

$\rm \longrightarrow \: \dfrac{dy}{dx} =y \bigg( {sec}^{2} x \: \: log \: {x} + \dfrac{tan \: x}{x} \bigg)$

Now , put y = x^tanx

$\rm \longrightarrow \: \dfrac{dy}{dx} = {x}^{tan \: x} \: \bigg( {sec}^{2} x \: \: log \: {x} + \dfrac{tan \: x}{x} \bigg)$

Answer :-

$\rm{x}^{tan \: x} \: \bigg( {sec}^{2} x \: \: log \: {x} + \dfrac{tan \: x}{x} \bigg)$

Additional Information :

d(e^x)/dx = e^x

d(x^n)/dx = n x^(n-1)

d(ln x)/dx = 1/x

d(sin x)/dx = cos x

d(cos x)/dx = - sin x

d(tan x)/dx = sec² x

d(sec x)/dx = tan x * sec x

d(cot x)/dx = - cosec²x

d(cosec x)/dx = - cosec x * cot x