Asking to simplify

Question

d/dx ∫₀^{5x^3 cos(t) dt}

^{Upper: 5x^3}

Lower: 0

Answer 1

This is an application of the fundamental theorem of Calculus.

The Fundamental Theorem tells us that F'(x) = f(x) when x is the upper bound of the integral.

Then, from the chain rule, F(g(x))' = F'(g(x))g'(x) =f(g(x))g'(x)

Then, for the above problem, we get cos(5x^3)(5x^3)' = cos(5x^3)(15x^2)

Answer 2

Let's start by evaluating the integral.

5x³

∫cos(t) dt = sin(5x³)-sin(0) = sin(5x³)

0

Then we take the derivative d/dx (sin(5x³)) = cos(5x³)*15x²

Where we have taken this derivative using the chain rule.

There is another way to do this problem. I recommend looking up the Leibniz Integral Rule, which tells you how you can "bring the d/dx into the integral".