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

Jan 11, 2017

$G ' \left(x\right) = 2 {x}^{5} - 6 {x}^{3}$

#### Explanation:

We have:

$G \left(x\right) = {\int}_{1}^{{x}^{2}} \left({t}^{2} - 3 t\right) \setminus \mathrm{dt}$

If we let

$F \left(x\right) = {\int}_{a}^{x} \left({t}^{2} - 3 t\right) \setminus \mathrm{dt}$

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

$F ' \left(x\right) = {x}^{2} - 3 x \text{ } \ldots \ldots . \left[1\right]$

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

${\int}_{a}^{x} \left({t}^{2} - 3 t\right) \setminus \mathrm{dt} = F \left(x\right) - F \left(a\right) \text{ } \ldots \ldots . \left[2\right]$

Therefore,

$G \left(x\right) = F \left({x}^{2}\right) - F \left(a\right) \text{ }$ from [2]

Differentiating wrt $x$ we get

$\setminus \setminus \frac{d}{\mathrm{dx}} G \left(x\right) = \frac{d}{\mathrm{dx}} F \left({x}^{2}\right) - \frac{d}{\mathrm{dx}} F \left(a\right)$
$\therefore G ' \left(x\right) = F ' \left({x}^{2}\right) \frac{d}{\mathrm{dx}} \left({x}^{2}\right) - 0 \text{ }$ chain rule
$\text{ "= ((x^2)^2-3(x^2)) (2x) " }$ from [1]
$\text{ } = \left({x}^{4} - 3 {x}^{2}\right) \left(2 x\right)$
$\text{ } = 2 {x}^{5} - 6 {x}^{3}$