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.
arc tan t
where t = x^2
so
arc tan x^2
(it is how fast the area is changing at the right side of the area, the high limit of t)
so after i did this problem i was stuck on two answer choices: arctan(x^2) or 2x arctan (x^2)
No problem! I can help you find the derivative of the given function h(x) using the Fundamental Theorem of Calculus.
The Fundamental Theorem of Calculus states that if F(x) is a function such that F'(x) = f(x), then the derivative of the integral of f(x) from a to x is equal to f(x). In other words, if we have an integral with a variable upper limit, we can differentiate it by simply evaluating the integrand at the variable upper limit and multiplying by the derivative of the variable.
Now, let's apply this principle to find the derivative of h(x) in your question.
Given: h(x) = ∫(2 to x^2)arctan(t) dt
To find h'(x), we need to evaluate the derivative of h(x) with respect to x. However, this integral has a variable upper limit, so we'll need to apply the Chain Rule.
Using the Chain Rule, we have:
h'(x) = d/dx [∫(2 to x^2)arctan(t) dt]
To differentiate the integral, we'll apply the Fundamental Theorem of Calculus and evaluate the integrand at the upper limit (x^2), then multiply by the derivative of the upper limit (x^2).
So, let's differentiate the integrand (arctan(t)):
d/dx [arctan(t)] = 1/(1+t^2) * dt/dx
Now, substitute t = x^2:
d/dx [arctan(x^2)] = 1/(1+(x^2)^2) * d(x^2)/dx
Simplifying further, we have:
= 1/(1+x^4) * 2x
= 2x/(1+x^4)
Finally, substitute this result back into our original equation:
h'(x) = 2x/(1+x^4)
Therefore, the derivative of h(x) is h'(x) = 2x/(1+x^4).