A function is defined by $$f(x,y) = \int_{x^2}^{xy} e^{t^2} \,dt$$ We are to decide the partial derivatives of the function.
I am quite unsure whether this is a trick question because we could do an approximation of the integrand with the Taylor series, but I can't see how that would simplify the work.
Could I just get a direction to as how I should approach this problem? I would very much not like to just get the solution.
Note that this is one of the easiest questions on the paper.
Hint: If we let $F(s) = \displaystyle\int_{0}^{s}e^{t^2}\,dt$, then $f(x,y) = F(xy)-F(x^2)$.
Now, try using the chain rule to compute $\dfrac{\partial f}{\partial x} = \dfrac{\partial }{\partial x}\left[F(xy)-F(x^2)\right]$ and $\dfrac{\partial f}{\partial y} = \dfrac{\partial }{\partial y}\left[F(xy)-F(x^2)\right]$. Remember that by construction, $F'(s) = e^{s^2}$.
\begin{align} & f(x,y) = \int_{x^2}^{xy} e^{t^2} \,dt = \int_0^{xy} e^{t^2} \, dt - \int_0^{x^2} e^{t^2} \, dt. \\[10pt] & \frac d {du} \int_0^u e^{t^2} \, dt = e^{u^2}. \\[10pt] & \frac d {dx} \int_0^u e^{t^2}\,dt = \frac d {du} \int_0^u e^{t^2} \, dt \times \frac{du}{dx} \end{align}
You have $$f(x,y)=\int_{a(x,y)}^{b(x,y)} g(t) \, dt$$ Using the fundamental theorem of calculus $$\frac{\partial f(x,y)}{\partial x}=\frac{\partial b(x,y)}{\partial x}\, g(b(x,y))-\frac{\partial a(x,y)}{\partial x}\, g(a(x,y))$$ $$\frac{\partial f(x,y)}{\partial y}=\frac{\partial b(x,y)}{\partial y}\, g(b(x,y))-\frac{\partial a(x,y)}{\partial y}\, g(a(x,y))$$
Just apply.
So basically the fundamental theorem of calculus gives us $$f(x,y) = \int_{x^2}^{xy} e^{t^2} \,dt = \int_{0}^{xy} e^{t^2} \,dt - \int_{0}^{x^2} e^{t^2} \,dt$$ $$\implies$$ $$\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x} (F(xy) - F(x^2)) =yF'(xy) - 2xF'(x^2) = ye^{x^2 y^2}-2xe^{x^4}$$
$$\dfrac{\partial f}{\partial y} = xe^{x^2y^2}$$
Thanks for the help, I knew this was supposed to be easy. Just looking at it from a different perspective!
