- #1

mathmari

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Let $I=[a,b]$, $J=[c,d]$ compact intervals in $\mathbb{R}$, $g,h:I\rightarrow J$ differentiable, $fI\times J\rightarrow \mathbb{R}$ continuous and partial differentiable as for the first variable with continuous partial derivative.

Let $F:I\rightarrow \mathbb{R}$.

I want to calculate the derivative of $$F(x)=\int_{g(x)}^{h(x)}f(x,y)\, dy$$ using the chain rule.

Following hint is given:

Let $G:I\rightarrow \mathbb{R}^3$ and $H:J\times J\times I\rightarrow \mathbb{R}$ defined by $G(x):=(g(x), h(x), x)=(u,v,w)$ and $H(u,v,w):=\int_u^vf(w,y)\, dy$. Then it is $F=H\circ G$. So, to calculate the derivative of the integral we have to calculate the derivative of $F(x)=H(G(x))$.

From the chain rule we have that $F'(x)=H'(G(x))\cdot G'(x)$.

The derivatives of the functions $H$ and $G$ are the following:

\begin{align*}G'(x)&=\frac{dG(x)}{dx}=\frac{dG(x)}{dg(x)}\cdot \frac{dg(x)}{dx}+\frac{dG(x)}{dh(x)}\cdot \frac{dh(x)}{dx}+\frac{dG(x)}{x}\cdot \frac{dx}{dx}\\ & =\frac{dG(x)}{dg(x)}\cdot \frac{dg(x)}{dx}+\frac{dG(x)}{dh(x)}\cdot \frac{dh(x)}{dx}+\frac{dG(x)}{x}\end{align*} The last term $\frac{dG(x)}{x}$ is not the same as $G'(x)$, is it?

For the derivative of $H$ do we use the total differential?

Or am I thinking wrong? (Wondering)