Question

How do you find the derivative of `G(x)=int (t^2-3t)dt` from `[1,x^2]`?

Answer 1

`G'(x) = 2x^5-6x^3`

Explanation:

We have:

`G(x) = int_1^(x^2) (t^2-3t) \ dt`

If we let

`F(x) = int_a^x (t^2-3t) \ dt`

Then from the first part of the Fundamental Theorem of Calculus, we have:

`F'(x) = x^2-3x " "....... [1]`

And from the second part of the Fundamental Theorem of Calculus, we have:

`int_a^x (t^2-3t) \ dt = F(x) - F(a) " "....... [2]`

Therefore,

`G(x) = F(x^2) - F(a) " "` from [2]

Differentiating wrt `x` we get

`\ \d/dx G(x) = d/dx F(x^2) - d/dx F(a)`
`:. G'(x) = F'(x^2)d/dx (x^2) - 0 " "` chain rule
`" "= ((x^2)^2-3(x^2)) (2x) " "` from [1]
`" "= (x^4-3x^2) (2x)`
`" "= 2x^5-6x^3`