Today I learnt about implicit differentiation using this:

$\frac{d}{dx}(f)$ = $\frac{df}{dy} \times \frac{dy}{dx}$

I don't understand when doing implicit differentiation how the d/dy part works for y terms:

$\frac{d(y^2)}{dx} = \frac{d(y^2)}{dy}\times\frac{dy}{dx}=2y\frac{dy}{dx}$

Why is the derivative of $y^2$ with respect to $y$, $2y$? Do you assume it's a function of something else? And similarly if you apply this same formula to an $x$ term (even though its not needed) you get:

$\frac{d(x^2)}{dx} =\frac{d(x^2)}{dy}\times\frac{dy}{dx}$

How does that simplify to the $2x$ I know it is?


  • $\begingroup$ What is $y$ in your last differentiation? Take $y=x$ then it works perfectly. $\endgroup$ – A.Γ. Nov 26 '18 at 18:14
  • $\begingroup$ @A.Γ. Its just an example, I'm not really sure, I thought of it as $y=x^2$ and $y=f$and did it using the formula at the top. $\endgroup$ – DevinJC Nov 26 '18 at 18:16
  • $\begingroup$ If $y=x^2$ it still works perfectly. $\endgroup$ – A.Γ. Nov 26 '18 at 18:18

Yes, here they assume that $y$ is a function of something else.

Whenever we apply implicit differentiation like with $\frac{dy^2(x)}{dx}$, we sort of first forget it is a function of something else, say $x$, we just derive it using the chain rule and then we need to correct for this via the familiar additional factor:

$$ \frac{d y^2(x)}{d x} =2 y(x) \cdot\frac{d y(x) }{dx}$$

Actually implict differentiation often boils down to just using the chain rule. Notice that if $y$ is not a function of $x$, we get that $\frac{ d y}{dx} =0$ and the equation just reads $0=0$.

For your second case where you state: $$\frac{d(x^2)}{dx} =\frac{d(x^2)}{dy}\times\frac{dy}{dx}$$ We could call $y=x^2$ because you really need to tell the reader what this new variable $y$ is in relation to the old variable $x$, else the derivative could be anything. We then get: $$ \frac{dy}{dx} =\frac{dy}{dy}\times\frac{d x^2 }{dx}= 1 \cdot 2x $$

In general we use implicit differentiation when there is some complicated variable dependence and you cannot directly express some variable explicitly in terms of the other. An example of such an equation would be: $$t+ t^2 + 3z^2 + z^3 = 6$$ Using implicit differentiation we could construct the tangent line for $(t,z)=(1,1)$, we will derive both sides with respect to $t$, we do not know if $z$ is is a function of $t$ so we need to use the chain rule.

We would get: $$1+ 2t + 6z \frac{dz}{dt} + 3 z^2 \frac{dz}{dt} = 0$$ Now we plug in the value at (1,1) to get: $$1+ 2 + 6 \frac{dz}{dt} + 3 \frac{dz}{dt} = 0$$ Or after rearranging: $$9 \frac{dz}{dt} = -3 \rightarrow \frac{dz}{dt}= -\frac{1}{3}$$ We get the tangent line $z= -\frac{1}{3}(t-1) +1$

  • $\begingroup$ Thanks a lot this really helped. $\endgroup$ – DevinJC Nov 26 '18 at 18:35
  • $\begingroup$ any time ;) we're all here to help $\endgroup$ – Wesley Strik Nov 26 '18 at 18:36
  • $\begingroup$ Hello, I just saw your edit. why dont the $\frac{dy}{dx}$ terms cancel out for the y terms as the $ \frac{dx}{dy}$ terms for the x terms. Or do we just not do that otherwise we can't rearrange for what we want ($\frac{dy/dx}$) $\endgroup$ – DevinJC Nov 27 '18 at 10:10
  • $\begingroup$ Maybe I should use different letters to make clear that $y$ is not always equal to $x^2$ $\endgroup$ – Wesley Strik Nov 27 '18 at 12:53
  • 1
    $\begingroup$ It automatically IS, because the function is a relation between $x$ and $y$, but we only derive with respect to $x$. If you would use the chain rule like this for $X=x$: $$ \frac{d}{dx} (X(x)^2)= 2X \cdot \frac{dx}{dx}=2X=2x$$ I think you get what you mean by using the chain rule for a variable you don't need to use it for. Here I mean that $X$ is a function of $x$ (clearly). $\endgroup$ – Wesley Strik Nov 27 '18 at 14:30

You implicitly utilize the formula of derivative of a composite functions or a chain rule: $$ f = y(x)^2, \qquad\frac{df}{dx}=\frac{d(y^2)}{dx},\\ f=g(y(x)),\qquad g(y)=y^2,\\ \frac{d}{dx}g(y(x))=\frac{dg(y(x))}{dy}\frac{dy(x)}{dx}=(2y(x))\frac{dy}{dx} $$


The derivative of $y^2$ with respect to $y$ is $2y$ because of the definition of derivative: $$\lim_{h \to 0} \frac{(y+h)^2-y^2}{h} = 2y.$$

In the last expression you write $$\frac{d(x^2)}{dx} = \frac{d(x^2)}{dy}\frac{dy}{dx}$$ but this is meaningful only when $x$ is a function of $y$. So assume $x = f(y)$. Then, by the first equation you mention in the question, the right hand side is $$\frac{d(f^2)}{dy}\frac{dy}{dx} = 2f(y)f'(y)\frac{dy}{dx} = 2xf'(y)\frac{dy}{dx} = 2x\frac{dx}{dy}\frac{dy}{dx} = 2x\frac{dx}{dx} = 2x.$$


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