0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
8
$\begingroup$

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 $$H'(x)=2x\sin^3(x^2)$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.