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$\def\d{\mathrm{d}}$We know that it is true that$$\lim_{\Delta x \to 0} \frac{f(x + \Delta x) -f(x)}{\Delta x} = \frac{f(x + \d x) -f(x)}{\d x} = f'(x),$$ where $\d x$ is define to be an infinitesimal.

Then we could rearrange the equation and say that$$f(x + \d x) -f(x) = f'(x) \,\d x.$$

Will this last equation be valid or correct?

Update. Do you agree that:

$$f'(x) \,\d x = \int_{x}^{x+\d x}f'(x)\,\d x.$$

Is this last equation valid or making sense? Does it even mean anything if you put a $dt$ in the limit? What I meant by valid is that would it be possible to apply it like in the context of the question here

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    $\begingroup$ What is an infinitely small increment? Are you talking about infinitesimals and hyperreals? $\endgroup$ – user251257 Apr 29 '17 at 11:28
  • $\begingroup$ It is the linear approximation formula. Google it. $\endgroup$ – StubbornAtom Apr 30 '17 at 4:26
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Yes its valid we use this result in cases where we want to find an approximate value without using calculator. For eg say we want the value of $\sqrt {64.1} $ so we define $f (x)=\sqrt {x}$ then using your equation its $\sqrt {64.1} \approx \sqrt {64}+0.1 \frac {1}{2\sqrt {64}}=8+0.1/16=8.00625$

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  • $\begingroup$ Does it look fine now? $\endgroup$ – Archis Welankar Apr 29 '17 at 11:27
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In Questions involving these kind of tricks, Taylor series expansion is always useful and very powerful
From Taylor Series Expansion of $f(x + dx)$ about $x=x$,
\begin{equation} f(x+dx)=f(x)+f^{'}(x)dx+f^{''}(x)(\frac{(dx)^{2}}{2!}) +f^{'''}(x)(\frac{(dx)^{3}}{3!}) + .... \end{equation} If $f(x)$ is defined and all derivatives are defined at $x=x$ then we can approximate above equation by neglecting higher order terms under following assumption that our step-size $dx$ tends to $0$.
And we have following relation \begin{equation} f(x+dx)\approx f(x)+f^{'}(x)dx \end{equation} \begin{equation} f(x+dx)- f(x) \approx f^{'}(x)dx \end{equation}
This is just an approximation though, not exact equal sign because you have neglected higher order terms.
This limiting taylor series expansion has numerous applications especially in Computational-science and numerical methods for solving ordinary differential equations.
(Sorry, I don't have good latex skills though :) )

Update: To answer your question in update, I don't see error in that equation but again it should have approximate sign and the derivative should be defined

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  • $\begingroup$ It says here that the question is flawed. $\endgroup$ – Gin99 Apr 30 '17 at 0:23
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The equation is only valid where the differential exists.

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The equation $f(x+dx)-f(x)=f'(x)\>dx$ is a valid identity in the variables $x$ and $dx$ only in the case when $f(x)=ax+b$ for certain constants $a$ and $b$. But this is not what you have in mind.

On the other hand, if $f$ is differentiable at the point $x$ then $$f(x+dx)-f(x)=f'(x)\>dx+o(dx)\qquad(dx\to0)\ .$$ This is a true and precise statement following immediately from the definition of the derivative. It means that in the approximate equation $$f(x+dx)-f(x)\approx f'(x)\>dx\qquad(dx\to0)$$ the error is of essentially smaller order of magnitude than the term $f'(x)\>dx$ on the right hand side (at least, when $f'(x)\ne0$).

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