Chain Rule for derivatives: $\frac{d}{dx} \left( \int_{a(x)}^{b(x)} g(x, y) dy \right) $

Question

Let $$f(x) = \int_{a(x)}^{b(x)} g(x, y) dy $$

I want to find $f'(x_0) $ using chain rule.

Solution

Let \begin{align*}\mathbb{R} \stackrel{h_1}{\to} \mathbb{R}^3 \stackrel{h_2}{\to} \mathbb{R}\end{align*}

where $h_1(x) = (x, x, x)$ and $$h_2(x_1, x_2, x_3) = \int_{a(x_2)}^{b(x_1)} g(x_3, y) dy$$

Then, $f = h_2 \circ h_1$ \begin{align*} f'(x_0) &= h_2'(h_1(x_0)) . h_1'(x_0)\\ &= \begin{bmatrix} \frac{\partial h_2 }{\partial x_1}\Big|_{(x_0, x_0, x_0)} & \frac{\partial h_2 }{\partial x_2}\Big|_{(x_0, x_0, x_0)} & \frac{\partial h_2 }{\partial x_3}\Big|_{(x_0, x_0, x_0)} \end{bmatrix} \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}\\ \\ &= \frac{\partial h_2 }{\partial x_1}\Big|_{(x_0, x_0, x_0)} + \frac{\partial h_2 }{\partial x_2}\Big|_{(x_0, x_0, x_0)} + \frac{\partial h_2 }{\partial x_3}\Big|_{(x_0, x_0, x_0)} \tag{1} \end{align*}

Lastly, I have:

$$\frac{\partial h_2 }{\partial x_1}\Big|_{(x_0, x_0, x_0)} = g(x_0, b(x_0)).b'(x_0) \tag{2}$$

$$\frac{\partial h_2 }{\partial x_2}\Big|_{(x_0, x_0, x_0)} = -g(x_0, a(x_0)).a'(x_0) \tag{3} $$

$$\frac{\partial h_2 }{\partial x_3}\Big|_{(x_0, x_0, x_0)} = g(x_0, b(x_0)) - g(x_0, a(x_0)) \tag{4}$$

Plugging this back in $(1)$ I believe is the final solution.

I want to verify this solution especially $(2)$, $(3)$ and $(4)$ and get corrections (if any).

Answer 1

$\frac{\partial h_2}{\partial x_3}$ should be $$\int_{a(x_2)}^{b(x_1)}\frac{\partial g}{\partial x_3}(x_3,y)dy$$. Thus the final answer is $$f'(x_0)=g(x_0,b(x_0))b'(x_0)-g(x_0,a(x_0))a'(x_0)$$ $$+\int_{a(x_0)}^{b(x_0)}\frac{\partial g}{\partial x}(x_0,y)dy$$