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Let $L(x) = \int_{1}^{x}f(t)dt$, can I say that $\frac{\partial L(x)}{\partial f(x)} = \frac{f(x)}{f'(x)}$?

Proof) $\frac{\partial L(x)}{\partial f(x)} = \frac{\partial L(x)}{\partial x}\frac{\partial x}{\partial f(x)} = \frac{f(x)}{f'(x)}$

shashack
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    No. ${}{}{}{}{}$ – copper.hat Jan 30 '22 at 06:51
  • Is there anwer form for the derivative of $\frac{\partial L(x)}{\partial f(x)}$? Could you tell me what I am wrong for the proof? – shashack Jan 30 '22 at 06:56
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    There is a first notational issue: $L(x) = \int_{1}^{x}f(x)dx$ should be $L(x) = \int_{1}^{x}f(t)dt$ (you cannot have the same letter for a bound and for the integration variable). – Jean Marie Jan 30 '22 at 07:48
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    What is wrong is that it's a formal manipulation, but how do you define the derivation with respect to a function ? There is a framework in which - perhaps - this computation (derivation with respect to a function) could make sense, it's the so-called "calculus of variations" – Jean Marie Jan 30 '22 at 07:53
  • Thank for pointing out that. I just changed dx to dt :) – shashack Jan 30 '22 at 07:58
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    If $f$ and $g$ are both differentiable at $x$, $f'(x)\neq 0$, then you may define $$ \frac{{\partial g(x)}}{{\partial f(x)}} = \mathop {\lim }\limits_{h \to 0} \frac{{g(x + h) - g(x)}}{{f(x + h) - f(x)}}. $$ But this would be just $g'(x)/f'(x)$. – Gary Jan 30 '22 at 08:12

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Yeah, it's true but there are some other conditions to be mentioned. Let $f$ be any continuous and differentiable function in the interval $[1,x]$ for some $x\in\mathbb{R}$. And $f'(v)\neq 0$ for $v\in[1,x]$. Now denote, $$L(x)=\int_{1}^{x}f(u) du$$ By the fundamental theorem of calculus we have that, $$L'(x)=f(x)$$ $$\frac{dL}{dx}=f(x)$$ Similarly we have that, $$f'(x)=\frac{df}{dx}$$ Their ratio gives, $$\frac{dL}{df}=\frac{f(x)}{f'(x)}$$ As required.

RAHUL
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  • You should assume $f'(x)\neq 0$. How do you define $dL/df$? – Gary Jan 30 '22 at 08:12
  • You don't need $f'>0$, $f'<0$ is also fine. You need $f'(x)\neq 0$ for that particular $x$. – Gary Jan 30 '22 at 08:14
  • I define it as ratio between two infinitesimals @Gary. As we are working in field of reals we have that limit definition of derivatives is same thing as $\frac{df}{dx}$. It might change strictly if we work with hyperreals (Model theory and non standard analysis and such stuff) – RAHUL Jan 30 '22 at 08:16
  • I understand $df(x)/dx$, but I am asking for the definition of $dL/df$. – Gary Jan 30 '22 at 08:17
  • You need $f'(x)$ for that particular $x$ not for the whole interval. – Gary Jan 30 '22 at 08:18
  • I define it as ratio between $\frac{dL}{dx}$ and $\frac{df}{dx}$. – RAHUL Jan 30 '22 at 08:18
  • I see. But there is almost nothing to prove here. – Gary Jan 30 '22 at 08:20
  • @Gary, Since we are defining $L(x)$ I think we need $f'(v)$ in that entire interval. – RAHUL Jan 30 '22 at 08:20
  • It seemed to me that you are differentiating at a particular $x$. – Gary Jan 30 '22 at 08:21
  • To define $L(x)$ I need f to be defined in that entire interval. But to clarify op's question we are required to consider f at x too. – RAHUL Jan 30 '22 at 08:22
  • But $f$ (and $L$) being defined on that interval has nothing to do with $f' \neq 0$ on that interval. The only point where $f' \neq 0$ is needed is the last step, where $f'(x)\neq 0$ is needed for that particular $x$. – Gary Jan 30 '22 at 08:25
  • The presence of $f'(x)$ in the denominator reminds me of a useful formula: $\delta(f(x))=\textstyle \sum_i\frac{\delta(x-a_i)}{\lvert f^\prime(a_i)\rvert}$ as given here. – Jean Marie Jan 30 '22 at 09:15