Partial term in Leibniz rule

Question

I'm having difficulties grasping why Leibniz rule the way it is: $$ \frac{d}{dx} \int^{b(x)}_{a(x)}f(x,t)dt = f(x,b(x))\frac{db}{dx} - f(x, a(x))\frac{da}{dx} + \int^{b(x)}_{a(x)}\frac{\partial}{\partial x}f(x,t)dt $$

Specifically, I don't understand how the last integral term comes to be? i.e. $$ \int^{b(x)}_{a(x)}\frac{\partial}{\partial x}f(x,t)dt $$

I think I understand the source of the first 2 terms:

Let $\frac{dF(x)}{dx} = f(x)$, then $$ \int^{b(x)}_{a(x)}f(x,t)dt = F(x,b(x)) - F(x,a(x)) $$ differentiating: $$ \frac{d}{dx}[F(x,b(x)) - F(x, a(x))] = f(x, b(x))\frac{db}{dx} - f(x, a(x))\frac{da}{dx} $$ $\frac{db}{dx}$ and $\frac{da}{dx}$ come from Chain Rule

Where does $$ \int^{b(x)}_{a(x)}\frac{\partial}{\partial x}f(x,t)dt $$ come from, what am I missing?

Answer 1

Note that $f$ and $F$ are bivariate functions and that $F$ corresponds to the antiderivative of $f$ with respect to the second argument only, i.e. $f(x,y) = \partial_yF(x,y)$. When dealing with several variables, it is often easier to work with differentials. Here, one has : $$ \mathrm{d}F(x,y) = \frac{\partial F}{\partial x}\mathrm{d}x + \frac{\partial F}{\partial y}\mathrm{d}y = \partial_xF(x,y)\mathrm{d}x + f(x,y)\mathrm{d}y, $$ hence $$ \frac{\mathrm{d}}{\mathrm{d}x} \left(F(x,b(x))-F(x,a(x))\right) = \left[\partial_xF(x,y) + f(x,y)\frac{\mathrm{d}y}{\mathrm{d}x}\right]_{y=a(x)}^{y=b(x)} = \partial_xF(x,b(x)) - \partial_xF(x,a(x)) + f(x,b(x))b'(x) - f(x,a(x))a'(x), $$ where $$ \partial_x\left(F(x,b(x))-F(x,a(x))\right) = \partial_x \int_{a(x)}^{b(x)} f(x,y) \,\mathrm{d}y = \int_{a(x)}^{b(x)} \partial_xf(x,y) \,\mathrm{d}y. $$ In conclusion, this last term comes from the partial derivative of $f$ with respect to its first argument. It is to be noted that this partial derivative commutes with the integral because the $x$-dependent bounds of the integral come into play through its second argument only, while the total derivative doesn't, because it takes into account the contributions coming from both arguments (hence the said Leibniz integral rule).

Answer 2

$$F(x):=\int^{b(x)}_{a(x)}f(x,t)dt=G(x,a(x),b(x)),$$ where $$G(x,u,v)=\int_u^vf(x,t)dt.$$ Under reasonable regularity hypothesis, $$\partial_vG(x,u,v)=f(x,v),\quad\partial_uG(x,u,v)=-f(x,u),\quad\partial_xG(x,u,v)=\int_u^v\partial_xf(x,t)dt$$ and by the chain rule, $$F'(x)=f(x,b(x))b'(x)-f(x,a(x))a'(x)+\int^{b(x)}_{a(x)}\partial_xf(x,t)dt.$$

Answer 3

If you read carefully your proof, you will notice that sometimes your $F$ depends on $t$ and sometimes it does not. Your proof is not correct. As suggested by José, you can check out https://en.wikipedia.org/wiki/Leibniz_integral_rule#Proofs