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???
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)
To find the derivative of the function h(x) = ∫(from 2 to x^2)arctan(t) dt, you can use the second fundamental theorem of calculus, also known as the Leibniz Rule.
The Leibniz Rule states that if the upper limit of the integral is a function of x (in this case, x^2), then the derivative of the function with respect to x is given by:
h'(x) = f(x^2) * (d/dx) x^2
Let's break down the steps to find h'(x):
1. Find the antiderivative of the integrand, arctan(t), with respect to t. The antiderivative of arctan(t) is simply t * arctan(t).
So, the integral becomes: h(x) = ∫(from 2 to x^2) t * arctan(t) dt
2. Apply the Leibniz Rule: h'(x) = [(x^2) * arctan(x^2)] * (d/dx) x^2
3. Simplify the expression by taking the derivative of x^2 with respect to x: (d/dx) x^2 = 2x
So, h'(x) = 2x * [(x^2) * arctan(x^2)]
Therefore, the correct answer is 2x * arctan(x^2).
Note: The first fundamental theorem of calculus states that if F(x) is the antiderivative of f(x), then the integral of f(x) from a to b is equal to F(b) - F(a). However, in this case, since the upper limit of the integral depends on x, we need to use the Leibniz Rule instead.