Find derivative of sin^5 (3x-2)?

1 Answer
Mar 17, 2018

#(dy)/(dx)=15sin^4(3x-2)cos(3x-2)#

Explanation:

#y=sin^5(3x-2)=(sin(3x-2))^5#, take,#u=sin(3x-2)#
#=>y=u^5 and u=sin(color(red)(3)x-2)#
#=>(dy)/(du)=5u^4 and (du)/(dx)=cos(3x-2)*color(red)(3)#
We have a chain rule,
#color(blue)((dy)/(dx)=(dy)/(du).(du)/(dx))#
#:.(dy)/(dx)=5u^4cos(3x-2)*3#
#=>(dy)/(dx)=15u^4cos(3x-2)#,where, #u=sin(3x-2)#
#=>(dy)/(dx)=15sin^4(3x-2)cos(3x-2)#
Hint :
#(1)d/(dX)(X^n)=n*X^(n-1)#
#(2)d/(dX)(sinX)=cosX##and d/(dx)(sin(3x-2))=cos(3x-2)d/(dx)(3x-2)=cos(3x-2)*3#