Let $F\left(x\right)=\int_{t^2}^{x^3}{\frac{\mathrm{d}y}{\sqrt{x^2+y^4}}}$. Find $F'\left(x\right)$

Question

Let $$F\left(x\right)=\int_{t^2}^{x^3}{\frac{\mathrm{d}y}{\sqrt{x^2+y^4}}}$$ Find $F'\left(x\right)$

Had the lower limit been a function of $x$ I would have easily calculated. I am not very familiar with the Newton-Leibniz rule in two variables. I do not know which formula to use.

Answer 1

You can use the most general one. The below is from Wikipedia.

So actually because the lower limit of the integral is a function of $t$ and because you are differentiating with respect to $x$, you can treat $t^2$ as a constant. And you can change the integrating variable, and for the purpose of preventing confusion let's change it to $t'$. Notice that the $t$ inside the integral is independent of the $t^2$ in the lower limit.

$$F_t(x) = \int_{a(t)}^{b(x)} g(x, t') \ dt'$$

$$\frac{\partial F_t(x)}{\partial x} = g(x, b(x)) \times \frac{\partial b}{\partial x} \ + \ g(x, a(t)) \times \frac{\partial a(t)}{\partial x} + \int_{a(t)}^{b(x)} \frac{\partial g(x, t')}{\partial x} dt'$$

If we plug in everything we get

$$f_t(x) = \frac{3x^2}{\sqrt{x^2 + x^6}} - \ 0 \ -\frac{1}{2} \int_{t^2}^{x^3} (t'^4 + x^2)^{-3/2} dt'$$

I couldn't integrate the last term, so you may try it out.

What's important here is to think about the intuition behind the formula - think of it as a proper generalisation and combining of the two formulae:

1. $$\frac{d}{dx} \int_a^x f(t) dt = f(x) \Rightarrow \frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x)) \times \frac{db}{dx} - f(a(x)) \times \frac{da}{dx}$$ (which is a simple use of chain rule)

2. $$\frac{d}{dx} \int_a^b f(x, t) \ dt = \int_a^b \frac{\partial f(x, t)}{\partial x} \ dt$$

Answer 2

A simple solution to our problem would be either removing $ x $ from the integrand, or removing it from the bounds, in general it is not always possible to remove the $ x $ from the integrand, it's always possible to remove it from the bounds though.

Since we're dealing with : $ F_{t}: x\mapsto\int_{t^{2}}^{x^{3}}{\frac{\mathrm{d}y}{\sqrt{x^{2}+y^{4}}}} $, then, for $ \left(x,t\right)\in \mathbb{R}^{2} $, using the change of variable $ \left\lbrace\begin{aligned}y&=\left(x^{3}-t^{2}\right)u+t^{2}\\ \mathrm{d}y&=\left(x^{3}-t^{2}\right)\mathrm{d}u\end{aligned}\right. $, we're left with : $$ F\left(x\right)=\left(x^{3}-t^{2}\right)\int_{0}^{1}{\frac{\mathrm{d}u}{\sqrt{x^{2}+\left(\left(x^{3}-t^{2}\right)u+t^{2}\right)^{4}}}} $$

It's now possible to use the product rule, then to switch the derivative with the integral sign in order to calculate our derivative : \begin{aligned} \small\frac{\mathrm{d}F}{\mathrm{d}x}\left(x\right)&=\small 3x^{2}\int_{0}^{1}{\frac{\mathrm{d}u}{\sqrt{x^{2}+\left(\left(x^{3}-t^{2}\right)u+t^{2}\right)^{4}}}}+\left(x^{3}-t^{2}\right)\int_{0}^{1}{\frac{\partial}{\partial x}\left(\left(x,u\right)\mapsto\frac{1}{\sqrt{x^{2}+\left(\left(x^{3}-t^{2}\right)u+t^{2}\right)^{4}}}\right)\,\mathrm{d}u} \\ &=\small 3x^{2}\int_{0}^{1}{\frac{\mathrm{d}u}{\sqrt{x^{2}+\left(\left(x^{3}-t^{2}\right)u+t^{2}\right)^{4}}}}-\left(x^{3}-t^{2}\right)\int_{0}^{1}{\frac{12u x^{2}\left(\left(x^{3}-t^{2}\right)u+t^{2}\right)^{3}+2x}{2\left(x^{2}+\left(\left(x^{3}-t^{2}\right)u+t^{2}\right)^{4}\right)^{\frac{3}{2}}}\,\mathrm{d}u}\\ \frac{\mathrm{d}F}{\mathrm{d}x}\left(x\right)&=\frac{3x^{2}}{x^{3}-t^{2}}\int_{t^{2}}^{x^{3}}{\frac{\mathrm{d}y}{\sqrt{x^{2}+y^{4}}}}-\frac{1}{x^{3}-t^{2}}\int_{t^{2}}^{x^{3}}{\frac{6\left(y-t^{2}\right)x^{2}y^{3}+\left(x^{3}-t^{2}\right)}{\left(x^{2}+y^{4}\right)\sqrt{x^{2}+y^{4}}}\,\mathrm{d}y}\end{aligned}