How do you differentiate #f(x)=(sin^3x^2))^(3/2)# using the chain rule?

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Answer 1 · 2015-12-03T00:40:24

#f'(x)=9xcos(x^2)sin^(7/2)(x^2)#

Recall that #sin^3x^2=(sinx^2)^3#.

Thus, we can simplify: #f(x)=(sinx^2)^(9/2)#

Through the chain rule:

#f'(x)=9/2(sinx^2)^(7/2)d/dx[sinx^2]#

Find the derivative:

#d/dx[sinx^2]=(cosx^2)d/dx[x^2]=2xcosx^2#

Thus:

#f'(x)=9xcos(x^2)sin^(7/2)(x^2)#