Question

Find dy/dx when y = x^sinx + tanx^cotx

Answer 1

Step-by-step explanation:

$y = {x}^{ \sin(x) } + { \tan(x) }^{ \cot(x) }$

◾taking log on both sides

$log(y) = log( {x}^{sinx} + {tanx}^{cotx} )$

$logy = (log \: {x}^{sinx} + log {tanx}^{cotx} )$

$logy = (sinx \: logx + cotx \: log(tanx))$

◾Differentiate wrt X

$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (sinx \: logx \: + cotx.log(tanx))$

$= //((sinx \frac{1}{x} + logx(cosx)) + (cotx \times \frac{1}{tanx} </p><p>\times {sec}^{2} x+logtanx\times (-cosec^2x))$

$= //( \frac{sinx}{x} + logx(cosx) </p><p>+ \frac{cot \times {sec}^{2}x }{tanx} +logtanx\times (-cosec^2x)$

$= //y( \frac{sin}{x} + logx(cosx)</p><p> + \frac{cotx \times {sec}^{2} x}{tanx}+logtanx\times (-cosec^2x) )$

$\frac{dy}{dx}=//{x}^{ \sin(x) } + { \tan(x) }^{ \cot(x) }(( \frac{sin}{x} + logx(cosx) + \frac{cotx \times {sec}^{2} x}{tanx}+logtanx\times (-cosec^2x) )$

Answer 2

Your answer attached in the photo