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

Nov 3, 2017

The Fundamental Theorem of Calculus states that, if $y \left(x\right) = {\int}_{a}^{x} f \left(t\right) \setminus \mathrm{dt}$, then $\frac{\mathrm{dy}}{\mathrm{dx}} = f \left(x\right)$.

The problem gives a function $G \left(x\right)$ defined as ${\int}_{1}^{{x}^{2}} {e}^{{t}^{2}} \setminus \mathrm{dt}$.

Since the bound is ${x}^{2}$, if we apply the Fundamental Theorem, we get $\frac{\mathrm{dG}}{d \left({x}^{2}\right)} = {e}^{{x}^{2}}$. However, we want to know what $\frac{\mathrm{dG}}{\mathrm{dx}}$ is.

To do this, we apply the chain rule. Given $\frac{\mathrm{dG}}{d \left({x}^{2}\right)}$, if we multiply this by $\frac{d \left({x}^{2}\right)}{\mathrm{dx}}$, we get $\frac{\mathrm{dG}}{\cancel{d \left({x}^{2}\right)}} \cdot \frac{\cancel{d \left({x}^{2}\right)}}{\mathrm{dx}} = \frac{\mathrm{dG}}{\mathrm{dx}}$.

Using the power rule, $\frac{d \left({x}^{2}\right)}{\mathrm{dx}} = 2 x$.

Thus, $\frac{\mathrm{dG}}{\mathrm{dx}} = \frac{\mathrm{dG}}{d \left({x}^{2}\right)} \cdot \frac{d \left({x}^{2}\right)}{\mathrm{dx}} = 2 x {e}^{{x}^{2}}$