Calculate $\frac{\partial^{2}z}{\partial x\partial y}$ for $z(x,y)=\int_{1}^{x^{2}-y^{2}}[\int_{0}^{u}\sin(t^{2})dt]du.$

Let $$z(x,y)=\int_{1}^{x^{2}-y^{2}}[\int_{0}^{u}\sin(t^{2})dt]du.$$ Calculate $$\frac{\partial^{2}z}{\partial x\partial y}$$

I tried to solve this using the Fundamental Theorem of Calculus.

I also found an solution like this: using Fundamental Theorem of Calculus, we get: $$\frac{\partial z}{\partial y}=\left[\int_{0}^{x^{2}-y^{2}}\sin(t^{2})dt\right]\cdot(-2y)$$ I can't understand why the extremes have changed and why I have to multiply by the partial derivate of $y$ in $x^{2}-y^{2}$.

Thanks.

• If $F(x) = \int^{x^2}_1 f(t) \, dt$, do you understand why $F'(x) = f(x^2)\, 2x$? If so, apply the same reasoning to the above question. Edit - the reasons are the FTC, and the chain rule... 2nd Edit: re-reading your question, it seems to me you must have understood this, no? – peter a g Nov 28 '15 at 3:28

As I noted in the comment section, I am a bit puzzled at your confusion, since you seem to arrive at $\partial z / \partial y$ - however, maybe this will help....
Write $$F( y ) = \int_1^{x^2 -y^2} f ( u ) \,du,$$ where one views $x$ as a constant. Then, by the FTC and the chain rule, one has $$F'(y) = f( x^2 - y^2)\cdot (-2y).$$ In this question, $$f(u) = \int^u_0 \sin ( t^2) \, dt,$$ so, substituting in, one has $$F'(y) =\int^{x^2-y^2}_0 \sin ( t^2) \, dt \cdot (-2y),$$
which is your calculated expression for $\partial z/ \partial y$.
Note that $z$ is the composite map $$z: (x,y) \mapsto x^{2}-y^{2} =: \xi \mapsto \int_{1}^{\xi}\int_{0}^{u}\sin t^{2} dt du;$$ so by chain rule and fundamental theorem of calculus we first have $$D_{2}z(x,y) = \int_{0}^{\xi}\sin t^{2}dt\cdot (-2y) = \int_{0}^{x^{2}-y^{2}}\sin t^{2}dt \cdot (-2y),$$ and then we have $$D_{1,2}z(x,y) = (-2y)\cdot \sin(x^{2}-y^{2})^{2}\cdot (2x)$$ again by chain rule and fundamental theorem of calculus.