Question

The derivative of log (sin(logx) (x > 0) ​jfssss

Answer 1

Explanation:

Explanation:

Explanation:

Answer:

( cos ( ㏒ x ) ) / ( x . sin ( ㏒ x ) )

Step-by-step explanation:

Let :

y = ㏒ ( sin ( ㏒ x ) )   ,   x > 0

# ㏒ x = ㏒_e x = ㏑ x

We know :

= > ( ㏑ x )' = 1 / x

= > y' = 1 / sin ( ㏒ x ) . ( sin ( ㏒ x ) )'

= >  y' = 1 / sin ( ㏒ x ) . ( cos ( ㏒ x ) ) . ( ㏒ x )'

= > y' = 1 / sin ( ㏒ x ) . ( cos ( ㏒ x ) ) . 1 / x

= > y'  = ( cos ( ㏒ x ) ) / ( x . sin ( ㏒ x ) )

Hence we get required answer!

Answer 2

$Let, \: y = log \: \sin( logx) \\ \\ \ = > \frac{dy}{dx} = \frac{d}{dx} (log \sin( logx)) \\ \\ = \frac{d. log \sin( logx)}{ \: d. \sin logx} . \frac{d. \sin (logx)}{dx} \\ \\ = \frac{1}{ \sin( logx) }. \frac{d. \sin( logx)}{d. logx } . \frac{d. logx}{dx} \\ \\ = \frac{1}{ \sin( logx) } . \cos( logx ). \frac{1}{x} \\ \\ = \frac{ \cot( logx) }{x} \\ \\ = \frac{1}{x} \cot( logx)$

I hope this helps you.