h(x) = ∫[2,x^2] arctan(t) dt
from the 2nd Fundamental Theorem of calculus (and the chain rule),
h'(x) = arctan(x^2) * 2x
if F(t) = ∫arctan(t), then
h(x) = F(x^2) - F(2)
now the chain rule gives the result, and d/dx F(2) = 0
If we had had
h(x) = ∫[lnx,x^2] arctan(t) dt
then we'd have ended up with
h'(x) = 2x arctan(x^2) - 1/x arctan(lnx)
let the function h(x)= (integrand symbol from 2 to x^2)arctan (t) dt. Find h'(x).
This question confused me because i know the derivative of an integral is the original function. I just need help with finding the derivative of that integral. I don't know how to post integral symbol sorry about that.
so after i did this problem i was stuck on two answer choices: arctan(x^2) or 2x arctan (x^2)
I don't which is the answer because i used the first fundamental theorem of calc and i somehow ended up with both???
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