1
$\begingroup$

Say there is an equation $y=x^2$. I know if we differentiate this equation we get $$\frac{\mathrm dy}{\mathrm dx}·1=2x·\frac{\mathrm dx}{\mathrm dx}\Longrightarrow\frac{\mathrm dy}{\mathrm dx}=2x.$$

The definition of $\dfrac{\mathrm dy}{\mathrm dx}$ is differentiating $y$ with respect to $x$, and $\dfrac{\mathrm dx}{\mathrm dx}$ is differentiating $x$ with respect to $x$. I understand $x^2$ becomes $2x$ by differentiating $x$ with respect to $x$, but I do not understand the $y$ part. Shouldn't it become $0$ from $y$? Shouldn't we consider $y$ as constant since we are not differentiating with respect with $y$?

$\endgroup$
3
  • $\begingroup$ In your first line you define $y$ as a function of $x$, ie. $y(x) = x^{2}$ so it's not constant upon differentiation w.r.t $x$. Do you know the definition of the differential operator $\frac{d}{dx}$? $\endgroup$ Jun 23 '18 at 3:08
  • $\begingroup$ I think d/dx only means to differentiate $\endgroup$
    – user547075
    Jun 23 '18 at 3:24
  • $\begingroup$ Yes, but I mean, do you know what is differentiation doing? If you don't know what I mean, you can read about the definition of the derivative here $\endgroup$ Jun 23 '18 at 3:28
5
$\begingroup$

Shouldn't it become 0 from y? Shouldn't we consider y as constant since we are not differentiating with respect with y?

Let's say $y=x^2$.   That is, $y$ is a dependent variable.   Such is not constant; nor could it be "held constant with respect to $x$" for a partial derivative (which this is not).   So therefore its derivative with respect to $x$ is clearly not zero (except when evaluated at $x=0$).   We evaluate it by substitution.$$\begin{align}\dfrac{\mathrm d ~~}{ \mathrm d x}y & =\dfrac{\mathrm d ~~}{\mathrm dx}x^2 \\[1ex]&= 2x\dfrac{\mathrm d ~~}{\mathrm dx}x\\[1ex]&=2x\end{align}$$

$\endgroup$
1
  • $\begingroup$ Oh that's what I was looking for ! Thanks a lot ^^ $\endgroup$
    – user547075
    Jun 23 '18 at 3:40
2
$\begingroup$

Note that $\frac {dx}{dx} =\frac {d}{dx}(x)$ means derivative of $ y=x$ with respect to $x$.

When you differentiate $y=x$ with respect to $x$ you get $\frac {d}{dx}(x) =1$

Usually we do not include $\frac {dx}{dx}$ when differentiating $ y=x^2$.

We simply write $\frac {dy}{dx}= 2x$

$\endgroup$
8
  • $\begingroup$ That is correct, you do not need to include it in your calculations. $\endgroup$ Jun 23 '18 at 3:18
  • $\begingroup$ But isn't (dy/dy)*y=1? So I thought dy/dx*y must be different since it is not differentiating with respect to y. $\endgroup$
    – user547075
    Jun 23 '18 at 3:22
  • $\begingroup$ d/dx* y=d/dx* x^2 so since we're differentiating with respect to x on both sides, I think the way they are being differentiated should be different since they are two different things $\endgroup$
    – user547075
    Jun 23 '18 at 3:29
  • $\begingroup$ Note that (dy/dy)=1 and (dy/dy)*y=y . dy/dx*y does not simplify unless you know what is y. $\endgroup$ Jun 23 '18 at 3:31
  • $\begingroup$ Oh yeah I meant d/dy * y =1 ^^ $\endgroup$
    – user547075
    Jun 23 '18 at 3:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.