Question

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

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

$\huge\boxed{\fcolorbox{cyan}{red}{Solution}}$

Let y=sin(logx)

⇒ dx=dy

dx=d [sin(logx)]=cos(logx)

dx=d (logx)= xcos(logx)

∴ dx =2d

2 y = dx

d [ xcos(logx) ] = x 2x.

dx.d [cos(logx)]−cos(logx)

dx

d (x)= x 2

x[−sin(logx).

dx=d (logx)]−cos(logx).1= x 2−xsin(logx). x1 −cos(logx)

= x 2−[sin(logx)+cos(logx)]