Question

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

Answer 1

Explanation:

Explanation:

( 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

Answer:

y = sin(log x)

dy/dn = d/dx .(sin log x) = cos (log x) . d/dx . (log x)

= cos (log x) . 1/x . d/dx (x)

= (cos (log x))/ x . 1

= (cos (log x))/x