# Fundamental Theorem of Calc Problem using Chain Rule

1. Sep 9, 2016

### ManicPIxie

1. The problem statement, all variables and given/known data
F(x) = (integral from 1 to x^3) (t^2 - 10)/(t + 1) dt
Evaluate F'(x)

2. Relevant equations
Using the chain rule

3. The attempt at a solution
Let u = x^3
Then:
[((x^3)^2 - 10) / (x^3 + 1)] ⋅ 3x^2
*step cancelling powers of x from fraction*
= (x^3 - 10)(3x^2)
= 3x^5 - 30x^2

I am trying to input a solution to a Maple TA problem for uni work, is this answer incorrect, or is possible that my maple TA syntax is incorrect.
My attempted answer on maple looked like:
(3*x^5) - (30*x^2)

Any help that *doesn't* just give me the answer is very much appreciated!

2. Sep 9, 2016

### ShayanJ

This is correct. But what happened to the denominator after this?

Last edited: Sep 9, 2016
3. Sep 9, 2016

### ManicPIxie

In the cancelling step:
[(3x^6 - 10)/(x^3 + 1)] ⋅ 3x^2
It cancelled to give:
(x^3 - 10) ⋅ 3x^2

4. Sep 9, 2016

### ShayanJ

You can't do that!
I don't even see how you're making that mistake to explain why its wrong!

5. Sep 9, 2016

### ManicPIxie

Retry:
[(x^6 - 10)/(x^3 + 1)] ⋅ 3x^2
= (3x^8 - 30x^2)/(x^3 + 1)

6. Sep 9, 2016

### ManicPIxie

Aha! And entering that into Maple TA works!