I am attempting to differentiate $y=e^{-x^2}\int_0^xe^{t^2}dt+c_1e^{-x^2}$ wrt x
My solution manual says that the answer is
$y'=e^{-x^2}e^{x^2}-2xe^{-x^2}\int_0^xe^{t^2}dt-2c_1xe^{-x^2}$
When I initially performed this operation myself, I wound up with
$y'=-2xe^{-x^2}\int_0^xe^{t^2}dt-2c_1xe^{-x^2}$
treating the $\int_0^xe^{t^2}dt$ as a constant.
Can anyone clarify for me where the $e^{-x^2}e^{x^2}$ came from? There must be a product rule that I am just missing to have that third term involved right?
Edit: Taking Josh's suggestion, if I treat $\int_0^xe^{t^2}dt$ as another x-term, as I rightly should have, then use a product rule, I should end up at the solution manual's step.
Differentiating $e^{-x^2}\int_0^xe^{t^2}dt$ wrt x
$(\frac{d}{dx}e^{-x^2})(\int_0^xe^{t^2}dt)+(\frac{d}{dx}\int_0^xe^{t^2}dt)(e^{-x^2})$
Now I'm stuck on trying to integrate this in so that I can differentiate it in terms of x.
$\int_0^xe^{t^2}dt$
Also, thank you to Shaun, I did not know that MathJax worked in the title
$$y=e^{-x^2}\int_0^xe^{t^2}dt+c_1e^{-x^2}$$
We have \begin{align} y' &= \left(\frac{d}{dx}e^{-x^2}\right)\left(\int_0^xe^{t^2}\,dt\right)+\left(\frac{d}{dx}\int_0^xe^{t^2}\,dt\right)\left(e^{-x^2}\right) + c_1 \frac{d}{dx} \exp(-x^2) \\ &= \left(-2xe^{-x^2}\right)\left(\int_0^xe^{t^2}\,dt\right)+\left(e^{x^2}\right)\left(e^{-x^2}\right) -2x c_1 \exp(-x^2) \\ &= \left(e^{x^2}\right)\left(e^{-x^2}\right)-2xe^{-x^2}\int_0^xe^{t^2}\,dt -2x c_1 \exp(-x^2)\\ &= 1-2xe^{-x^2}\int_0^xe^{t^2}\,dt -2 c_1 x \exp(-x^2) \\ &= 1-2x\exp(-x^2) \left(\int_0^xe^{t^2}\, dt + c_1 \right) \end{align}
Notice that the solution manual didn't simplify $e^{-x^2}e^{x^2}=1$.