# What is the reason $\frac{\mathrm dy}{\mathrm dx}$ is computed the way it is being computed?

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 $y$?

• 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 $\frac{d}{dx}$? Jun 23 '18 at 3:08
• I think d/dx only means to differentiate Jun 23 '18 at 3:24
• 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 Jun 23 '18 at 3:28

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}

• Oh that's what I was looking for ! Thanks a lot ^^ Jun 23 '18 at 3:40

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 $\frac {dy}{dx}= 2x$

• That is correct, you do not need to include it in your calculations. Jun 23 '18 at 3:18
• 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. Jun 23 '18 at 3:22
• 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 Jun 23 '18 at 3:29
• Note that (dy/dy)=1 and (dy/dy)*y=y . dy/dx*y does not simplify unless you know what is y. Jun 23 '18 at 3:31
• Oh yeah I meant d/dy * y =1 ^^ Jun 23 '18 at 3:32