0
$\begingroup$

I know fundamental theorem of calculus states that $$\frac d {dx} \int^x_a f(t)dt=f(x)~.$$

But does $$\frac d {dx} \int^{f(x)}_a f(t)dt=f(f(x))~?$$ Or does it equal to $$\frac d {dx} \int^{f(x)}_a f(t)dt=f(f(x))f'(x)~?$$

$\endgroup$
3
  • $\begingroup$ This is not FTC. This is Leibniz's rule. $\endgroup$ – Allawonder May 9 '20 at 14:51
  • 1
    $\begingroup$ @Allawonder You're wrong. $\endgroup$ – Ted Shifrin May 9 '20 at 22:51
  • $\begingroup$ @TedShifrin No. You're wrong. $\endgroup$ – Allawonder May 9 '20 at 23:52
4
$\begingroup$

With $F(x):=\int_a^xf(t)\,\mathrm dt$, we have $F'(x)=f(x)$ and by the chain rule $$\frac{\mathrm d}{\mathrm dx} \int_a^{f(x)}f(t)\,\mathrm dt=\frac{\mathrm d}{\mathrm dx}F(f(x))=f'(x)F'(f(x))=f'(x)f(f(x)).$$

$\endgroup$
0
$\begingroup$

Here I am going to provide you a basic idea for solving such a problem.

We all know that the (first) fundamental theorem of calculus is just the particular case of the Leibniz Integral Rule where $~a(x) = a~$, a constant, $~b(x) = x~$, and $~f(x, t) = f(t)~$.

So if you know the Leibniz Integral Rule (it is also known as Differentiation under the integral sign), then you can tackled all such problems.

Leibniz Integral Rule (Differentiation under the integral sign):

Let $f(x, t)$ be a function of $x$ and $t$ such that both $f(x, t)$ and its partial derivative $\frac{\partial f}{\partial x}$ are continuous in $t$ and $x$ in some region of the $(x, t)$-plane, including $a(x) ≤ t ≤ b(x)$, and $ x_0 ≤ x ≤ x_1$. Also suppose that the functions $a(x)$ and $b(x)$ are both continuous and both have continuous derivatives for $x_0 ≤ x ≤ x_1$. Then, for $x_0 ≤ x ≤ x_1$, $$\frac{d}{dx}\left(\int_{a(x)}^{b(x)} f(x,t) dt\right)=\int_{a(x)}^{b(x)} \frac{\partial }{\partial x}f(x,t) dt +f( x, b(x)) \frac{db}{dx}-f( x, a(x)) \frac{da}{dx}$$

${}$

Now for your problem, $~a(x)=a~$, $~b(x)=f(x)~$ and $~f(x, t) = f(t)~$. So by the above rule, \begin{equation}\frac d {dx} \int^{f(x)}_a f(t)dt=\int_{a}^{f(x)} \frac{\partial }{\partial x}f(t) dt +f(f(x)) \left(\frac{d}{dx}f(x)\right)-f(a) \frac{da}{dx}\\=0+f(f(x))f'(x)-0~~~~~~~~~~~~~~~~~~~\\=f(f(x))f'(x)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\end{equation} and therefore $$\boxed{\frac d {dx} \int^{f(x)}_a f(t)dt=f(f(x))f'(x)}$$

$\endgroup$
2
  • 1
    $\begingroup$ The Leibnitz integral rule is a consequence of the fundamental theorem of calculus. The latter is not a “particular case” of the former. The Wikipedia article that you link to says as much. $\endgroup$ – amd May 9 '20 at 20:01
  • $\begingroup$ This is like making someone memorize the quotient rule in order to understand how to take the derivative of $1/x$. Or having someone learn the concept of multiplication in order to solve $3+3$. $\endgroup$ – Ted Shifrin May 9 '20 at 23:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.