Question

How do you calculate the derivative of `int3(sin(t))^4 dt` from `[e^x,1]`?

Answer 1

Use the Fundamental Theorem of Calculus, Part 1 (after rewriting) and use the chain rule.

Explanation:

g(x) = `int_(e^x)^1 3sin^4t dt`

To use the fundamental theorem in one form, we must have the constant as the lower limit of integration, so we rewrite:

g(x) = `-int_1^(e^x) 3sin^4t dt`

With `u=e^x`, we have: `g(x) = h(u) = -int_1^(u) 3sin^4t dt`

Use the chain rule to find `g'(x) = h'(u) (du)/dx`.

By FTC 1, `h'(u) = 3sin^4 u`

We also have `(du)/dx = d/dx(e^x) = x^e`.

So,

`g'(x) = 3sin^4u (du)/dx = 3sin^4(e^x)* e^x`