### 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.

#### Exercises

- Suppose that $y$ is a function of $x$ and $$y^5-xy+x^5=1$$ Find $dy/dx$ at the point $x=1,y=0$.
- 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.

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##### Calculus Refresher

- Previous: Linear approximations: approximation by differentials
- Next: Related rates

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