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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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  • $\begingroup$ It is a little hard to read your integrand, however you should understand my answer. $\endgroup$ – Will Fisher Jul 25 '16 at 19:21
  • $\begingroup$ @gbox I reformatted the equation; did I do it right? $\endgroup$ – Matthew Leingang Jul 25 '16 at 19:23
  • $\begingroup$ @MatthewLeingang edited $\endgroup$ – gbox Jul 25 '16 at 19:24
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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 $$H'(x)=2x\sin^3(x^2)$$

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  • $\begingroup$ Why do we help the chain rule? $\endgroup$ – gbox Jul 25 '16 at 19:27
  • $\begingroup$ Let $$f(x)=\int_3^x\sin^3(t)dt$$ then we have that $H(x)=f(x^2)$ hence we use the chain rule. $\endgroup$ – Will Fisher Jul 25 '16 at 19:30
  • $\begingroup$ Did that answer your question? @gbox $\endgroup$ – Will Fisher Jul 25 '16 at 19:53

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