Let $f:\mathbb{R}^2\rightarrow\mathbb{R}$ be a continuous function.Define $h:\mathbb{R}\rightarrow\mathbb{R}$ by, $$h(x)=\int_0^x\int_0^xf(u,v)dudv$$ What will be the value of $h'(1)?$
I am getting no clue about this problem.I tried but don't know how to approach.
Let $h(x)$ be given by the integral
$$h(x)=\int_0^x \int_0^x f(u,v)\,du\,dv$$
Then, using Leibniz's Rule for differentiating under the integral sign, the deriviative $h'(x)$ is given by
$$\bbox[5px,border:2px solid #C0A000]{h'(x)=\int_0^x f(u,x)\,du+\int_0^x f(x,v)\,dv} \tag 1$$
To see this perhaps a bit more easily, we let $F(x,v)\equiv \int_0^x f(u,v)\,du$. Then, we can write
$$h(x)=\int_0^x F(x,v)\,dv$$
Now, using Leibniz's Rule, we see that
$$h'(x)=F(x,x)+\int_0^x \frac{\partial F(x,v)}{\partial x}\,dv \tag 2$$
Using $F(x,x)=\int_0^x f(u,x)\,du$ and $\frac{\partial F(x,v)}{\partial x}=f(x,v)$ in $(2)$ reveals that
$$\bbox[5px,border:2px solid #C0A000]{h'(x)=\int_0^x f(u,x)\,du+\int_0^x f(x,v)\,dv}$$
which agrees with the result in $(1)$.
