The differential of a function $y(x)$ is defined as,

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

I didn't know that a differential is actually defined by the above equation and is a function of both $x$ and $dx$, but does (1) the motivation comes from the linearization of a function?

(2) Is it mathematically valid to multiply and divide by differentials?

$$\frac{dy}{dx} = \frac{dy}{dx}, \text{then} \; dy = \frac{dy}{dx}dx$$

If it's not, then can we do,

$$\Delta y/\Delta x = \Delta y/\Delta x, \text{then} \;\Delta y = \frac{\Delta y}{\Delta x} \Delta x $$

And take the limit as $\Delta x \rightarrow 0, \Delta y \rightarrow 0$? So the end result is the same as if we just multiplied by differentials?

(3) Can a differential be defined by $\Delta x = dx$ when $\Delta x \rightarrow 0$?

(4) Is there a difference between writing,

$$\frac{d}{dx} y \; \text{and} \; \frac{dy}{dx}$$

Where the first is a derivative and the second is a ratio of partials?


2 Answers 2


$\displaystyle\frac{dy}{dx}=lim_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}$. Hence, $\displaystyle\frac{dy}{dx}$ is a notation for this limit. Do not take it as a division.

Regarding (4), both are same but the second is not a division as I said.


Yes, the differential is the best linear approximation of a function in a neighborhood of a point $a$. To be more specific, we would like to find a linear function $df_a$ such that $f(a+\Delta x)-f(a) \approx df_a (\Delta x) $, i.e. the difference in $f$ will be linearly related to the change in $x$.

To be more precise, the differential of a function $f(x)$ at the point $a$ is a linear function $df_a (\Delta x) $ such that: $$f(a+\Delta x)=f(a)+ df_a(\Delta x) + o(\Delta x)_{\Delta x\to 0} $$ The differential is unique and from this definition and the definition of the derivative, $df_a(\Delta x) = f'(a)\cdot \Delta x$. Following this, $\frac{df_a (\Delta x)}{\Delta x} = f'(a)$, and another way of thinking about it is $\frac{dy}{dx}=f'(x)$.

  • $\begingroup$ @DWade64 I hope this answered your questions, but if not feel free to ask again what bothers you and I will be happy to answer. $\endgroup$ May 1, 2015 at 17:22

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