let $H(x)=\int_{3}^{x^2} (\sin t)^3 dt$. Find $H'(x)$.
I understand that this question is related to the fundamental theorem of calculus, but how should I approach it?
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Sign up to join this communitylet $H(x)=\int_{3}^{x^2} (\sin t)^3 dt$. Find $H'(x)$.
I understand that this question is related to the fundamental theorem of calculus, but how should I approach it?
Notice that by the fundamental theorem of calculus we have that if $$f(x)=\int_a^xg(t)dt$$ Then $$f'(x)=g(x)$$ So here we just have to apply the chain rule because if $$f(x)=\int_a^{h(x)}g(t)dt$$ Then $$f'(x)=g(h(x))h'(x)$$ So your answer is $$H'(x)=2x\sin^3(x^2)$$