Suppose that $f:[0,1]\times [0,1]\rightarrow\mathbb{R}$ has continuous partial derivatives. Define $F:[0,1]\times [0,1]\rightarrow\mathbb{R}$ by $$F(u,v)=\int_{0}^{u} f(v,y) \ dy$$If $u=u(x)$ and $v=v(x)$, find an expression for $\frac{\partial F}{\partial x}$.

I recognised that this is can be solved using Leibniz Rule. So, $$F(u(x),v(x))=\int_{0}^{u} f(v,y) \ dy$$ $$\frac{\partial F}{\partial x}=\int_{0}^{u} \frac{\partial}{\partial x} f(v,y) \ dy \ \ \ \ \text{by Leibniz Rule}$$

I know I then need to apply FTC, however, I am unsure of how to simplify the integrand. Does the integrand equal $$\frac{\partial f}{\partial v}\frac{dv}{dx}?$$ I am unsure of how to apply the chain rule in this instance.


$f,\ F$ are function of variables $x,\ y$ and $u,\ v$ respectively.

So chain rule implies $$ F_x=F(u,v)_x=F_u u' + F_vv'\ \ast$$

First consider partial derivatives of $F$ : $$ F_u=_{FTC}f(v,u),\ f(v,y)_v=f_x(v,y) (v)_v,\ F_v= \int^u_0\ f_x(v,y)\ dy $$

so that by $\ast$, $$ F_x =f(v,u)u' + \int_0^u \ f_x (v,y)\ dy\ v' $$

  • $\begingroup$ But $y$ is not a function of $x$, so wouldn't $\frac{\partial}{\partial x}f(v,y)=\frac{\partial f}{\partial v}\frac{dv}{dx}?$ $\endgroup$ – user557493 Jun 20 '18 at 1:20
  • $\begingroup$ What is $F$ in OP ? $\endgroup$ – HK Lee Jun 20 '18 at 1:22
  • $\begingroup$ Sorry, I have fixed a typo in my question. It should say $F(u,v)=\int_{0}^{u} f(v,y) \ dy$. I wish to calculate $\frac{\partial}{\partial x}f(v,y).$ where $v=v(x)$ $\endgroup$ – user557493 Jun 20 '18 at 1:31
  • $\begingroup$ I see. $f(v,y)_x=f_x(v,y)v'(x)$. $\endgroup$ – HK Lee Jun 20 '18 at 1:35
  • $\begingroup$ I am not entirely sure of that notation. Am I incorrect? $\endgroup$ – user557493 Jun 20 '18 at 1:44

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