Derivative of definite integral, the chain rule for such a function.

Question

Orginal post

When you calculate the derivative of $$ \int_{u(x)}^{v(x)} f(t) d t $$ what you must realize is that this is actually a compositum function. You have 3 functions: >$F(u, v)=\int_u^v f(t) d t$ and the functions $u(x)$ and $v(x)$, giving you. $$ \int_{u(x)}^{v(x)} f(t) d t=F(u(x), v(x)) . $$ The chain rule for such a function is, $$ \begin{aligned} \frac{d}{d x} F(u(x), v(x)) & =\frac{d F}{d u} \frac{d u}{d x}+\frac{d F}{d v} \frac{d v}{d x} \\ & =-f(u(x)) \cdot u^{\prime}(x)+f(v(x)) \cdot v^{\prime}(x) \end{aligned} $$

I don't understand this. $$ \frac{d F}{d u}=-f(u(x)) $$

Answer 1

Write: $$\int_u^vf(t)\,\mathrm{d}t=-\int_v^uf(t)\,\mathrm{d}t$$And then it will become clearer.

Alternatively, mimic the proof of the FTC but just observe the differences pick up a negative sign: $$\int_{u+h}^vf(t)\,\mathrm{d}t-\int_u^vf(t)\,\mathrm{d}t=-\int_u^{u+h}f(t)\,\mathrm{d}t$$

Answer 2

I presume you already understand:

$$\frac{\partial F}{\partial v}=f(v(x))$$

because that is just the Fundamental Theorem of Calculus.

For the lower bound, notice that: $F(u,v)=-G(u,v)$ where $G(u,v)=\int_v^u f(t)dt$, so

$$\frac{\partial F}{\partial u}=-\frac{\partial G}{\partial u}=-f(u(x))$$

per the Fundamental Theorem of Calculus applied to $G$.

Answer 3

$$F(u,v)=\int_u^vf(t)dt=F(a,v)-F(a,u)$$ and $$\frac d{d y}F(a,y)=f(y),$$ hence $$\frac{\partial}{\partial u}F(u,v)=-\frac d{d u}F(a,u)=-f(u).$$

Answer 4

For such an example, it would be so, right? $$ \begin{aligned} f(x)= & \int_{-x^4}^{x^2+\sin x} \sin \left(e^y+\pi y^6\right) d y \\ \frac{d f}{d x}= & -\sin \left(e^{x^2+\sin x}+\pi\left(x^2+\sin x\right)^6\right)(2 x+\cos x)- \sin \left(e^{-x^4}+\pi\left(-x^4\right)^6\right)(4 x^3) \end{aligned} $$

Answer 5

$\qquad\begin{align} \dfrac{\partial F(u,v)}{\partial u} &= \lim_{h\to 0}\dfrac{F(u+h,v)-F(u,v)}{h}\\[2ex]&=\lim_{h\mapsto 0}\dfrac{\displaystyle\int_{u+h}^v f(t)\,\mathrm d t-\int_u^v f(t)\,\mathrm d t}{h}\\[2ex]&=\lim_{h\to 0}\dfrac{\displaystyle\int_{u+h}^u f(t)\,\mathrm d t}{h}\\[2ex]&=\dfrac{\displaystyle\lim_{h\to 0}\dfrac{\partial ~~}{\partial h}\int_{u+h}^u f(t)\,\mathrm d t }{1}\\[2ex]&=\lim_{h\to0}-f(u+h)\\[2ex]&= -f(u)\end{align}$