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While understanding the proof of the Leibniz's formula there are some steps that I don't really understand.

Theorem: Suppose that $f:G \subset \mathbb{R}^2 \longrightarrow \mathbb{R}$ is a continuous function. Consider the function

\begin{align} I:[c,d] &\longrightarrow\, \mathbb{R}\\ y\,\,\, &\longmapsto\, I(y) = \int_{a(y)}^{b(y)} f(x,y)\, dx \end{align}

If $\frac{\partial f}{\partial y}$ exists and is continuous over G, then I'(y) exists, then we can write:

\begin{align} I'(y) = \frac{d}{dy}\int_{a(y)}^{b(y)}f(x,y)dx = \int_{a(y)}^{b(y)}\frac{\partial}{\partial y}f(x,y)dx+\int_{a(y)}^{b(y)}\frac{db(y)}{dy}f(b(y),y)-\int_{a(y)}^{b(y)}\frac{da(y)}{dy}f(a(y),y) \end{align}

They start the proof with seeing I(y) as a compound function. Then they write:

$\frac{dI}{dy} = \frac{\partial I}{\partial a}\frac{da}{dy}+\frac{\partial I}{\partial b}\frac{db}{dy} + \frac{\partial I}{\partial y}$

I think they use the chain rule but on which part? I don't really see how they get that,

Thanks in advance!

Ayoub Rossi
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1 Answers1

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Please, rush right out and get yourself a new book, because the authors of this one do not seem to be trying to help you. :)

Write $$ I(y) = H(a(y), b(y), y) $$ where $$ H(u, v, w) = \int_u^v f(x, w)~ dx. $$

Now the chain rule says that $$ \frac{dI}{dy} = \frac{\partial H}{ \partial u} (a(y), b(y), y) \cdot \frac{da}{dy} + \frac{\partial H}{ \partial v} (a(y), b(y), y) \cdot \frac{db}{dy} + \frac{\partial H}{ \partial w} (a(y), b(y), y) \cdot 1. $$ Fortunately for us, the first two partials of $H$ are easy to evaluate by the fundamental theorem of calculus: $$ \frac{\partial H}{ \partial v} (u, v, w) = f(u, w)\\ \frac{\partial H}{ \partial u} (u, v, w) = -f(u, w). $$ The third partial involves a bit more work (and some assumptions about $f$ that I don't recall -- perhaps that it's bounded or something?) and turns out to say that $$ \frac{\partial H}{ \partial w} (u, v, w) = \int_u^v \frac{\partial f}{\partial y}(x, y) ~dx $$

Combining these together gives the required result.

By the way, it's really helpful, in cases like this, to write things like $$ D_1 H $$ to mean "the derivative of $H$ with respect to its first argument", rather than requiring that the first argument be called $u$, etc.

John Hughes
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  • Thank you for your well-detailed explanation maybe this should help: They say to see I(y) as a compound function: $[c,d] \rightarrow R^3\rightarrow R $and $ y \rightarrow (a(y),b(y),y) \rightarrow \int_{a}^{b}f(x,y)dx$ – Ayoub Rossi Jun 05 '17 at 17:25
  • Right -- their unnamed function is my $H$. :) – John Hughes Jun 05 '17 at 17:30
  • Thank you! Everything is clear now :) ( Maybe just one last question, why can we say that I(y) = H(a(y),b(y),y)? ) – Ayoub Rossi Jun 05 '17 at 17:30
  • By definition of H. –  Jun 05 '17 at 17:34
  • Write out $H(a(y), b(y), y)$ by substituting, in the definition of $H$: replace each occurrence of $u$ with $a(y)$; each occurrence of $v$ with $b(y)$, and each occurrence of $w$ with $y$. Compare the result to the formula for $I(y)$. – John Hughes Jun 05 '17 at 17:34
  • I see, we get exactly the same, thanks! – Ayoub Rossi Jun 05 '17 at 17:44
  • Can you explain how the two partials of H are easy to evaluate with the fundamental theorem of calculus? Thanks in advance! – Ayoub Rossi Aug 19 '17 at 11:47
  • Fix u and w. Then f becomes a function of one var and H becomes an integral of this function whose upper limit is the variable v in H. The second FTC then applies. – John Hughes Aug 19 '17 at 16:49
  • Now that I have a real keyboard: For the $v$ partial: Fix $u, w$. Define $g(x) = f(x, w)$ and $$G(v) = \int_u^v f(x, w) ~ dx = \int_u^v g(x)~gx.$$ Then 2nd FTC says that $G$ is differentiable and $G'(v) = g(v)$. But the left side is just $\frac{\partial H}{\partial v}$, and the right is $g(v) = f(v, w)$. For the $w$ partial, rewrite $H$ as $-\int_v^u f(x, w)~ dx$ and do the same trick. – John Hughes Aug 19 '17 at 21:20