Math Insight

Implicit differentiation


There is nothing ‘implicit’ about the differentiation we do here, it is quite ‘explicit’. The difference from earlier situations is that we have a function defined ‘implicitly’. What this means is that, instead of a clear-cut (if complicated) formula for the value of the function in terms of the input value, we only have a relation between the two. This is best illustrated by examples.

For example, suppose that $y$ is a function of $x$ and $$y^5-xy+x^5=1$$ and we are to find some useful expression for $dy/dx$. Notice that it is not likely that we'd be able to solve this equation for $y$ as a function of $x$ (nor vice-versa, either), so our previous methods do not obviously do anything here! But both sides of that equality are functions of $x$, and are equal, so their derivatives are equal, surely. That is, $$5y^4{dy\over dx}-1\cdot y-x{dy\over dx}+5x^4=0$$ Here the trick is that we can ‘take the derivative’ without knowing exactly what $y$ is as a function of $x$, but just using the rules for differentiation.

Specifically, to take the derivative of the term $y^5$, we view this as a composite function, obtained by applying the take-the-fifth-power function after applying the (not clearly known!) function $y$. Then use the chain rule!

Likewise, to differentiate the term $xy$, we use the product rule $${d\over dx}(x\cdot y)={dx\over dx}\cdot y+x\cdot{dy\over dx}= y+x\cdot{dy\over dx}$$ since, after all, $${dx\over dx}=1.$$

And the term $x^5$ is easy to differentiate, obtaining the $5x^4$. The other side of the equation, the function ‘$1$’, is constant, so its derivative is $0$. (The fact that this means that the left-hand side is also constant should not be mis-used: we need to use the very non-trivial looking expression we have for that constant function, there on the left-hand side of that equation!).

Now the amazing part is that this equation can be solved for $y'$, if we tolerate a formula involving not only $x$, but also $y$: first, regroup terms depending on whether they have a $y'$ or not: $$y'(5y^4-x)+(-y+5x^4)=0.$$ Then move the non-$y'$ terms to the other side $$y'(5y^4-x)=y-5x^4$$ and divide by the ‘coefficient’ of the $y'$: $$y'={y-5x^4\over 5y^4-x}$$

Yes, this is not as good as if there were a formula for $y'$ not needing the $y$. But, on the other hand, the initial situation we had did not present us with a formula for $y$ in terms of $x$, so it was necessary to lower our expectations.

Yes, if we are given a value of $x$ and told to find the corresponding $y'$, it would be impossible without luck or some additional information. For example, in the case we just looked at, if we were asked to find $y'$ when $x=1$ and $y=1$, it's easy: just plug these values into the formula for $y'$ in terms of both $x$ and $y$: when $x=1$ and $y=1$, the corresponding value of $y'$ is $$y'={1-5\cdot 1^4\over 5\cdot 1^4-1}=-4/4=-1$$

If, instead, we were asked to find $y$ and $y'$ when $x=1$, not knowing in advance that $y=1$ fits into the equation when $x=1$, we'd have to hope for some luck. First, we'd have to try to solve the original equation for $y$ with $x$ replace by its value $1$: solve $$y^5-y+1=1$$ By luck indeed, there is some cancellation, and the equation becomes $$y^5-y=0$$ By further luck, we can factor this ‘by hand’: it is $$0=y(y^4-1)=y(y^2-1)(y^2+1)=y(y-1)(y+1)(y^2+1)$$ So there are actually three real numbers which work as $y$ for $x=1$: the values $-1,0,+1$. There is no clear way to see which is ‘best’. But in any case, any one of these three values could be used as $y$ in substituting into the formula $$y'={y-5x^4\over 5y^4-x}$$ we obtained above.

Yes, there are really three solutions, three functions, etc.

Note that we could have used the Intermediate Value Theorem and/or Newton's Method to numerically solve the equation, even without too much luck. In ‘real life’ a person should be prepared to do such things.


  1. Suppose that $y$ is a function of $x$ and $$y^5-xy+x^5=1$$ Find $dy/dx$ at the point $x=1,y=0$.
  2. Suppose that $y$ is a function of $x$ and $$y^3-xy^2+x^2y+x^5=7$$ Find $dy/dx$ at the point $x=1,y=2$. Find ${ d^2\,y \over dx\,^2 }$ at that point.