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

Jun 7, 2017

You should get ${e}^{{x}^{2}}$.

You can use Leibniz's integral rule.

$\frac{\mathrm{dG} \left(x\right)}{\mathrm{dx}} \equiv \frac{d}{\mathrm{dx}} \left[{\int}_{g \left(x\right)}^{h \left(x\right)} f \left(x , t\right) \mathrm{dt}\right]$

$= {\int}_{g \left(x\right)}^{h \left(x\right)} \frac{\partial f}{\partial x} \mathrm{dt} + f \left(x , h \left(x\right)\right) \frac{\mathrm{dh}}{\mathrm{dx}} - f \left(x , g \left(x\right)\right) \frac{\mathrm{dg}}{\mathrm{dx}}$

In this case, $g \left(x\right) = 1$ and $h \left(x\right) = x$, so:

• $\frac{\mathrm{dg}}{\mathrm{dx}} = 0$
• $\frac{\mathrm{dh}}{\mathrm{dx}} = 1$
• $f \left(x , h \left(x\right)\right) = f \left(x , t = x\right) = {e}^{{x}^{2}}$
• $f \left(x , g \left(x\right)\right) = f \left(x , t = 1\right) = {e}^{{1}^{2}} = e$
• $\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} {\left[{e}^{{t}^{2}}\right]}_{x} = 0$

Therefore:

$\textcolor{b l u e}{\frac{\mathrm{dG}}{\mathrm{dx}}} = {\cancel{{\int}_{1}^{x} 0 \mathrm{dt}}}^{0} + {e}^{{x}^{2}} \cdot 1 - \cancel{e \cdot 0}$

$= \textcolor{b l u e}{{e}^{{x}^{2}}}$

Jun 7, 2017

$\frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{x} {e}^{{t}^{2}} \setminus \mathrm{dt} = {e}^{{x}^{2}}$

#### Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{a}^{x} \setminus f \left(t\right) \setminus \mathrm{dt} = f \left(x\right)$ for any constant $a$

(ie the derivative of an integral gives us the original function back).

$\frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{x} {e}^{{t}^{2}} \setminus \mathrm{dt}$
$\frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{x} {e}^{{t}^{2}} \setminus \mathrm{dt} = {e}^{{x}^{2}}$