The derivative of log (sin(logx) (x > 0) kkcs
Answers
Explanation:
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!
Explanation:
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!