Derivative of a double integral

Question

I have a question. How can I take the derivative of the following double integral?

$$\frac{d}{dx} \int_x^1\int_0^x tf(t)g(s)dtds$$

for $x\leq 1$?

Thanks.

Answer 1

Hint: The (first part of the) fundamental theorem of calculus tells you how to differentiate with respect to an endpoint of integration. Also, $\int_x^1 \dots = - \int_1^x \dots$. Then of course, with nested functions, it's time for the chain rule.

Answer 2

Let $h(s,x)=\left(\displaystyle\int_{0}^{x}tf(t)dt\right)g(s)$, and assume that $f$ and $g$ are sufficiently regular, then we are dealing with $-\dfrac{d}{dx}\displaystyle\int_{1}^{x}h(s,x)ds$.

Since $f$ and $g$ are regular, then \begin{align*} \dfrac{d}{dx}\int_{1}^{x}h(s,x)ds&=h(x,x)+\int_{1}^{x}\dfrac{\partial h}{\partial x}(s,x)ds\\ &=h(x,x)+\int_{1}^{x}xf(x)g(s)ds\\ &=h(x,x)+xf(x)\int_{1}^{x}g(s)ds. \end{align*}