How do you find the derivative of #sin^2x^2#?

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Answer 1 · 2016-01-10T16:10:53

# 4xsinx^2cosx^2 #

before starting note that #sin^2(x) = (sinx)^2#

this form makes it'easier' to apply the chain rule.

rewriting #sin^2x^2 = (sinx^2)^2 #

differentiating using the chain rule gives:

#2(sinx^2).d/dx(sinx^2) = 2(sinx^2)(cosx^2).d/dx(x^2) #

# = 2(sinx^2)cosx^2(2x) #

#rArr d/dx(sin^2x^2 ) = 4xsinx^2cosx^2 #