Question

Integral question

Answer 1

\[f(x)=\int\limits\limits_{-3}^{x^2}t^4dt\]

Answer 2

First do this: \[
g(y) = \int\limits\limits_{-3}^{y}t^4dt
\]

Answer 3

Hera, is the question asking you to find \(\large f'(x)\) by chance? :o

Answer 4

Then consider the relationship between \(g(y)\) and \(f(x)\)

Answer 5

It is zepdrix

Answer 6

That is \[
f(x) = g(x^2)
\]

Answer 7

This means, by chain rule: \[
f'(x) = g'(x^2)(x^2)'
\]

Answer 8

\[
g'(x^2) = g'(y)
\]

Answer 9

Do you get it yet?

Answer 10

AH! I was doing the derivative of the 2nd part wrong

Answer 11

In short \[
f(x)=\int\limits\limits_{a}^{g(x)}h(t)dt \implies f'(x) = h(g(x))g'(x)
\]

Answer 12

Awesome!

Answer 13

Did you understand how I got that though, using just the fundamental theorem of calculus and chain rule?

Answer 14

yeah I'm a little slow but you explained it really well

Answer 15

Another way to think about it:\[
f(x)=\int\limits\limits_{a}^{g(x)}h(t)dt \implies \frac{df}{dg} = h(g(x))\\
\frac{df}{dx} = \frac{df}{dg}\frac{dg}{dx} = h(g(x))g'(x)
\]