How do we go from $f'(x) = \frac{dy}{dx}$ to $dy = f'(x)dx$?

Asked Jul 22 '13 at 19:54

Active Jul 22 '13 at 20:27

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As far as I know, the derivative of $y$ is defined as:

$$f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}h = \frac{dy}{dx}$$

So $\frac{dy}{dx}$ is a limit, not a fraction of real numbers. I often read expressions like:

$$dy = f'(x)dx$$

How do we go from $f'(x) = \frac{dy}{dx}$ to $dy = f'(x)dx$? Clearly not multiplying both sides for $dx$.

The answer from the linked question says that:

\begin{align*} \frac{dy}{dx} &= \frac{dy}{du} \frac{du}{dx} \text{(chain rule)} \\ \frac{dy}{dx} &= \frac1{\frac{dx}{dy}} \text{(Inverse Function theorem)} \end{align*}

This can help me transforming $f'(x) = \frac{dy}{dx}$ to $\frac{dy}{du} = f'(x) \frac{dx}{du}$.

Then how can I proceed getting rid of $du$?

edited Apr 13 '17 at 12:21

Community

asked Jul 22 '13 at 19:54

user16538

1

What is $\lim_{x_1\to x} x_1 - x$? – Antonio Vargas Jul 22 '13 at 19:59
Check your definitions of $dx$. The derivative was historically introduced by Fermat as a measure of slope of continuous functions. – Torsten Hĕrculĕ Cärlemän Jul 22 '13 at 20:01
You definitions are at least strange. We always have $\lim_{x\to x_1}(x-x_1)=0$. If $f$ is continuous, $\lim_{x\to x_1}(f(x)-f(x_1))=0$. Thus, your symbols $dx,df$ make little sense. See what I link above. – Pedro Jul 22 '13 at 20:07
I strongly recommend reading the link in Peter Tamaroff's comment. – Emily Jul 22 '13 at 20:09
Putting a $d$ in front of something does not mean "take the limit of this particular thing," nor is $\lim \frac{\Delta f}{\Delta x}=\frac{\lim\Delta f}{\lim\Delta x}$ (as you have essentially written) valid. Basically, you are making up rules and definitions that make everything inconsistent and meaningless. – anon Jul 22 '13 at 20:12
I strongly recommend reading Arkamis' comment. – Kaster Jul 22 '13 at 20:28
I have rewritten the question trying to be more clear on what I cannot understand. – user16538 Jul 22 '13 at 20:28
See my response here. – Cameron Williams Jul 22 '13 at 20:37

How do we go from $f'(x) = \frac{dy}{dx}$ to $dy = f'(x)dx$?

0 Answers0