# Differentiate y= log (sin x ) w.r.t x .

Solution:  We have ,

$y= log (sinx)$

putting $u=sinx$ , we get

$y=log$ $u$ and $u=sinx$

$\therefore$                               $\frac{\mathrm{d}y }{\mathrm{d} u}=\frac{1}{u }$          and      $\frac{\mathrm{d}u }{\mathrm{d} x}=cosx.$

Now ,                         $\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} u}\times \frac{\mathrm{d} u}{\mathrm{d} x}$

$\Rightarrow$                             $\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{u}\times cosx =\frac{1}{sinx}\times cosx=cotx.$

Hence ,                    $\frac{\mathrm{d} {log(sinx)} }{\mathrm{d} x}=cotx.$

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