When introducing this concept, most courses overlook the best examples: the basic ones. Implicit differentiation is what allows you to find the derivatives for so many functions and even prove some general properties of derivatives that it's ridiculous to teach the concept without emphasizing them.\(
\newcommand\abs[1]{\left\vert {#1} \right\vert}

Implicit Differentiation?

In case you're walking into the post cold, here's the situation: Instead of knowing a function, we know some property of that function expressed in an equation, and we want to find its derivative. The trick is to take the derivative of both sides of that equation anyway. For every term with our function in it, chain rule will pop out the derivative of our function, and these will be the only places it shows up. Then you combine those like terms and solve for the derivative. As an example, suppose you want to know the derivative at a point on a circle of radius \(r\text{:}\)
x^2 + y^2 &= r^2 && \comment{equation of circle} \\
2x + 2y\deriv{y} &= 0 && \comment{derivative of both sides} \\
\deriv{y} &= \frac{-2x}{2y} = -\frac{x}{y} && \comment{solve for derivative}
A lot of insight can be gained in this process. Here, we discover that the derivative of a circle can be expressed very cleanly in terms of the point \((x, y),\) without referencing the radius of the circle. Also, if you had instead solved for one variable explicitly, you would have had to deal with the way more obnoxious \(y = \pm\sqrt{r^2 - x^2},\) and the clean result above would have been obfuscated.

The most common usages of implicit differentiation are for curves rather than functions, all those guys with \(x\)'s and \(y\)'s on the same side. This post is focusing below on two things that aren't that:

Product Rule Implies Quotient Rule

You can derive quotient rule by naming the quotient \(\frac{u}{v}\) as a function, say \(w,\) multiplying the denominator \(v\) over, then using implicit differentiation and the product rule. In symbols, using the prime notation (\(w'\)) for derivatives[1]:
\frac{u}{v} &= w \\
u &= wv &&\comment{multiply by $v$ on both sides} \\
\derivop{u} &= \derivop{wv} && \comment{derivative of both sides}\\
\deriv{u} &= \deriv{w} v + w\deriv{v} && \comment{product rule}\\
\deriv{w} &= \frac{\deriv{u} - w\deriv{v}}{v} && \comment{solve for derivative} \\
&= \frac{\deriv{u} - \frac{u}{v}\deriv{v}}{v} && \comment{plug in $w$ from first line} \\
&= \frac{\deriv{u} v - u\deriv{v}}{v^2} && \comment{multiply by $\frac{v}{v}$}
where those last two steps are just to make things match the quotient rule. Usually, they're less necessary or more obvious when you're in a specific example, as below.

Ratio of Polynomials

An example of this process:

y &= \frac{x^2 - 1}{x^2 + 1} \\
y(x^2 + 1) &= (x^2 - 1) && \comment{multiply by denominator}\\
\deriv{y}(x^2 + 1) + y\derivop{x^2 + 1} &= \derivop{x^2 - 1} && \comment{derivative, product rule}\\
\deriv{y}(x^2 + 1) + y(2x)\derivx &= 2x\derivx && \href{../derivatives-of-specific-functions/#polynomials}{\comment{power rule}}\\
\deriv{y} &= \frac{2x(1-y)}{x^2 + 1} && \comment{solve for $\deriv{y}$}
where we could plug in the function for \(y\) if we needed everything in terms of \(x,\) but we probably wouldn't.

Tangent Function

An example of this process, deriving \(
\derivop{\tan u}\text{:}\)

\frac{\sin u}{\cos u} &= \tan u && \href{../trigonometry-formula-sheet/#definitions}{\comment{definition of tangent}} \\
\sin u &= \tan u\cos u && \comment{multiply by denominator}\\
\derivop{\sin u} &= \derivop{\tan u}\cos u + \tan u\,\derivop{\cos u} && \comment{derivative, product rule}\\
(\cos u)\deriv{u} &= \derivop{\tan u}\cos u + \tan u(-\sin u)\deriv{u} && \href{../derivatives-of-specific-functions/#trigonometric}{\comment{derivative of sine, cosine}}\\
\derivop{\tan u} &= \frac{\cos u + \sin u\tan u}{\cos u}\deriv{u} && \comment{solve for $\derivop{\tan u}$} \\
&= (1 + \tan^2 u)\deriv{u} && \comment{distribute the division} \\
&= \sec^2 u\,\deriv{u} && \href{../trigonometry-formula-sheet/#pythagorean}{\comment{Pythagorean identity}}

Derivative and Inverse Implies Derivative of Inverse

Suppose you know a function \(\renewcommand\deriv[1]{{#1}'}
f,\) its derivative \(\deriv{f},\) and its inverse function \(g\) that satisfies \(f(g(x))=x,\) by definition<!TODO>. You want to find the derivative of the inverse \(\derivop{g}.\) Then use implicit differentiation on the definition for the inverse function:
\[ \begin{aligned}
f(g(x)) &= x && \comment{definition of inverse} \\
\derivop{f(g(x))} &= \derivop{x} && \comment{derivative} \\
\deriv{f}(g(x))\deriv{g}(x) &= \derivxone && \comment{chain rule} \\
\deriv{g}(x) &= \frac{\derivxone}{\deriv{f}(g(x))} && \comment{solve for $\deriv{g}(x)$}
\end{aligned} \]
Yes, this does actually work for every[1:1] inverse function. I particularly like that the original function isn't needed in the end, only its derivative and its inverse. You could memorize this formula if you wanted, but it's such a short process that I recommend going through the steps every time, especially since sometimes a different form of the inverse definition comes to mind. Below are demonstrations of this.

From Powers to Roots

An example of this process:

\sqrt[n]{x} &= y && \comment{name the inverse}\\
x &= y^n && \comment{cancel it with original}\\
\derivxone &= ny^{n-1}\deriv{y} && \comment{power rule}\\
\deriv{y} &= \frac{\derivxone}{ny^{n-1}} && \comment{solve for $\deriv{y}$}\\
&= \frac{1}{n}\sqrt[n]{x^{-n-1}}\derivx && \comment{plug in first line}

Compare to this formula.

From Exponents to Logarithms

An example of this process:

\ln x &= y\\
x &= e^y && \comment{exponential of both sides}\\
\derivxone &= e^y\deriv{y} && \comment{derivative of both sides}\\
\deriv{y} &= \frac{\derivxone}{e^y} && \comment{solve for $\deriv{y}$} \\
\derivop{\ln x} &= \frac{\derivxone}{x} && \comment{plug in first lines}

Between Angles and Ratios

An example of this process:

\arcsin x &= y\\
x &= \sin y && \comment{sine of both sides}\\
\derivxone &= \cos y\deriv{y} && \comment{derivative of both sides}\\
\deriv{y} &= \frac{\derivxone}{\cos y} && \comment{solve for $\deriv{y}$} \\
\derivop{\arcsin x} &= \frac{\derivxone}{\sqrt{1-x^2}} && \comment{plug in first lines}
where in the last step, we use the Pythagorean identity in a clever way: If \(\sin y = x,\) then we can plug that into the identity and solve for \(\cos y,\) getting \(\pm\sqrt{1-\sin^2 x}.\) Then we choose the positive root, since the range of \(\arcsin x\) was defined<!TODO> so that the derivative would be positive. In most courses, they'll have you draw a right triangle <!TODO: such as the one below> to illustrate how you're using the Pythagorean theorem.

This process works for all six inverse trigonometric functions, each time using a Pythagorean identity at the end to put everything in terms of \(x.\) You do have to be careful with signs — the derivative of every arc-trig function is defined to be positive, and the derivative of every arc-co-trig function is defined to be negative, so when \(\derivop{\arcsec x}\) and \(\derivop{\arccsc x}\) each get an \(x\) in the denominator, you have to take the absolute value to appropriately ignore the sign of the input.


  1. If you try to use this procedure for a function such as the reciprocal function \(g(x)=x^{-1},\) you'll run into trouble since it is its own inverse. In this case, \(f=g,\) so you can't use \(f'\) to find \(g'\). <!TODO: Determine the properties required for a function to be its own inverse, or dependent on its own inverse.> ↩︎ ↩︎