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I just finish learning the chain rule and am now learning Implicit Differentiation and I am wondering: why is it not possible to take the derivative of $\dfrac{dy^2}{dx}$? Why do we need to apply the chain rule and find $(\dfrac{dy^2}{dy})\cdot\dfrac{dy}{dx}$?

I understand that $y^2$ is a function and it is equal to $[y(x)]^2$, but basically I am having trouble understanding $\dfrac{dy^2}{dy}$ then $\dfrac{dy}{dx}$. Why do we find the derivative of ${y^2}$ with respect to $y$, and not $x$?

Marra
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user5826
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2 Answers2

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Note that, if we're working with a function $y=y(x)$, then $y^2$ is just the usual multiplication $y(x)\cdot y(x)=y(x)^2$. This is the same as composing the function $$ f(x)=x^2 $$ with $y(x)$, which gives $f(y(x))=y(x)^2$.

Now, if we're searching the derivative of $y^2$ with respect to $x$, we must use the Chain Rule and consider $y^2=f(y)$, which gives us $$ \dfrac{d(y(x)^2)}{dx}=\dfrac{d(f(y(x))}{dx}=\dfrac{df(y)}{dy}\cdot\dfrac{dy}{dx} $$

And since $\dfrac{df(y)}{dy}=\dfrac{d(y^2)}{dy}=2y$, you get the result. You should notice that since it's an implicit derivative, $\dfrac{dy}{dx}$ will depend on the value of $y(x)$ (that's why it's implicit).

For example, let's suppose that $y(x)$ is given implicitely by the equation $x^2+y(x)^2=1$ and we're trying to find the value of $\dfrac{dy}{dx}$ in a given $x$ point where $y(x)\neq 0$. We derive both sides of the equation, and use the argument given above (before we introduced the example) to get $$ \dfrac{d}{dx}(x^2+y(x)^2)=0 $$ then $$ \dfrac{d(x^2)}{dx}+\dfrac{d}{dx}(y(x)^2)=0 $$ $$ 2x+2y\dfrac{dy}{dx}=0 $$ This is the part we used the forementioned argument. Now we can get the value we want: $$ \dfrac{dy}{dx}=\dfrac{-2x}{2y} $$

I hope it is clear enough!

Marra
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  • excellent. getting there. I never saw d/dx (x^2) as x^2/dx, that helped. can you explain or expand the d/dx (y(x))^2 portion? thank you – user5826 Jul 24 '13 at 19:15
  • @Juan the part you mentioned was missing a "d", I corrected it now. – Marra Jul 24 '13 at 19:19
  • Also, notice that $\dfrac{d}{dx}(y(x)^2)$, $\dfrac{dy(x)^2}{dx}$ and $\dfrac{dy^2}{dx}$ are all meant to be the same thing. – Marra Jul 24 '13 at 19:20
  • why do we consider y^2=f(y) and not y^2=f(x)? – user5826 Jul 24 '13 at 19:42
  • You are correct; since $y=y(x)$ we should use $f=f(y)=f(y(x))$. Note that we can derive $f$ with respect to $y$ (obtaining $\dfrac{df}{dy}=y^2$ in this case) or with respect to $x$ (using the Chain Rule as I did above). I just omitted the $x$ in the first case you mentioned to make things easier, but it should be there. – Marra Jul 24 '13 at 19:45
  • Can y^2 be stated as y(y(x))? Or is this wrong? – user5826 Jul 24 '13 at 19:49
  • It is wrong; you are confusing the multiplication of functions with the composition of them. Did I wrote something like that in my answer? I can correct it. – Marra Jul 24 '13 at 19:50
  • You wrote: f(y(x))=y(x)^2. I am having trouble understanding that and: Now, if we're searching the derivative of y^2 with respect to x, we must use the Chain Rule and consider y^2=f(y) – user5826 Jul 24 '13 at 19:54
  • Just notice that, for any number $y$, we're working with the function $f(y)=y^2$. In the particular case where we substitute $y=y(x)$ (that is, the number $y$ now depends on the other number $x$ or equivalently, $y$ is a function of $x$), then we have $f(y)=f(y(x))=y(x)^2$. Notice now that $f$ depends on $x$ (since $f$ depends on $y$ and $y$ depends on $x$). Since $y=y(x)$, I can say "derivative of $y$ with respect to $x$" and "derivative of $y(x)$ with respect to $x$"; they both mean the same. Is it clear now? – Marra Jul 24 '13 at 19:59
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    "Notice now that f depends on x (since f depends on y and y depends on x)." Excellent! Thank you. – user5826 Jul 24 '13 at 20:35
  • I am glad to be of help :) – Marra Jul 24 '13 at 20:39
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Given a two fractions, $\frac{a}{b}$ and $\frac{b}{c}$, how do you calculate $\frac{a}{c}$?

Let's relabel $y^2$:

$$\require{cancel}\begin{align} u&=y^2 \\ \frac{d}{dx}y^2&=\frac{du}{dx}=\cancel{\frac{du}{dy}\frac{dy}{dx}} \end{align}$$

Take the derivative of $u$ with respect to $y$: $$\frac{du}{dy}=2y$$

Adriano
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rurouniwallace
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  • and whats the reasoning for dy/dx? – user5826 Jul 24 '13 at 19:00
  • @Juan We're calculating $\frac{du}{dx}$. It's fraction, like any other. If you need to calculate $\frac{a}{c}$ and you know $\frac{a}{b}$, how do you calculate $\frac{a}{c}$? – rurouniwallace Jul 24 '13 at 19:03
  • @ZettaSuro $\dfrac{du}{dx}$ is not a fraction; in a calculus course it should not be introduced as so. – Marra Jul 24 '13 at 19:11
  • @GustavoMarra It is indeed a fraction, and can be treated as one in this situation. It is the fraction between two differentials, both of which cancel, which is the basis of the chain rule. This can be seen here, where the derivative is written in Leibniz notation. The very definition of the derivative is a quotient. – rurouniwallace Jul 24 '13 at 19:14
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    It is as you said; a notation, and should be used only as so, not as ratios; $dx$ and $dy$ are not numbers or elements of a field; they are more correctly seen as forms or as references to measures; therefore they cannot be "cancelled". I understand that this is very common in engineering courses but it is a wrong practice in front of more formal approaches for the theory. You can find more about this in any Analisys book which covers integration; it's not Calculus material I guess. – Marra Jul 24 '13 at 19:18