How would I use the fact that integration is the inverse of differentiation, to find $∂F/∂x$ and
$∂F/∂y$ for the case where :
$F(x, y)$ = $ \int_{0}^{xy^2} {e^{t^2}}\ dt$
I initially tried to use the chain rule to work this out, but realised that that would only give $dF/dx$ which is different. Am I right in thinking that I can just use the fact that $\frac{\partial}{\partial{x}}$ = $\frac{\partial}{\partial{t}}$$\frac{\partial{t}}{\partial{x}}$, and then bring the $\frac{\partial}{\partial{t}}$ inside the integral ( which I think I would be allowed to do since it would hold the limits constant but I'm not exactly sure). Furthermore, how would you go about calculating $\frac{\partial{t}}{\partial{x}}$, would one just partially differentiate the limits, even though $t$ is just a dummy variable?
If anyone could shed some light on this it would greatly help my understanding. Thanks!
Best Answer
Calculating $\partial_x F(x,y)$ We want to evaluate $$\partial_x F(x, y) = \frac{d}{dx}\int_{0}^{xy^2} {e^{t^2}}\,\mathrm{d}t.$$
Note that in this case our integrand $f(t)=e^{t^2}$ does not depend on $x$. This trivially means that $\partial_x f(t)=0$. Observe that $a(x)=0$ and $b(x)=xy^2$, so $$\partial_x F(x, y)=e^{(xy^2)^2}\cdot y^2-e^{0^2}\cdot 0=y^2e^{(xy^2)^2}.$$
Calculating $\partial_y F(x,y)$ We want to evaluate $$\partial_y F(x, y) = \frac{d}{dy}\int_{0}^{xy^2} {e^{t^2}}\,\mathrm{d}t.$$
Note again that in this case our integrand $f(t)=e^{t^2}$ does not depend on $y$. This trivially means that $\partial_y f(t)=0$. Observe that $a(y)=0$ and $b(y)=xy^2$, so $$\partial_y F(x, y)=e^{(xy^2)^2}\cdot x2y-e^{0^2}\cdot 0=2xy \,e^{(xy^2)^2}.$$