Implicit Differentiation

Background

So far, you may only have differentiated functions written in the form $y=f\left(x\right)$. But some functions are better described by an equation involving $x$ and $y$. For example, $x^{2}+y^{2}=16$ describes the graph of a circle with center $\left(0,0\right)$ and radius 4, and is really the graph of two functions: $y=\pm {\sqrt {16-x^{2}}}$.

Sometimes, functions described by equations in $x$ and $y$ are too hard to solve for $y$, for example $x^{3}+y^{3}=6xy$. This equation really describes 3 different functions of x, whose graph is the curve:

We want to find derivatives of these functions without having to solve for $y$ explicitly. We can do this by implicit differentiation, in which we take the derivative of both sides of our equation with respect to $x$, and do some algebra steps to solve for $y'$ (or ${\dfrac {dy}{dx}}$ if you prefer), keeping in mind that $y$ is a function of $x$ throughout the equation.

Example 1

1. Find $y'$ if $\sin y-3x^{2}y=8$.

[Momentarily think of $y=f\left(x\right)$ and view the equation as $\sin \left(f\left(x\right)\right)-3x^{2}\cdot f\left(x\right)=8$ to realize that the $\sin \left(f\left(x\right)\right)$ term requires the chain rule and the $3x^{2}\cdot f\left(x\right)$ term needs the product rule when we differentiate with respect to $x$, while the derivative of 8 is just 0.]

So we get \[ \begin{array}{rcl} \sin y-3x^{2}y & = & 8\\ \left(\cos y\right)\cdot y'-\left(3x^{2}y'+6xy\right) & = & 0\quad\textrm{(derivative of both sides with respect to \ensuremath{x})}\\ \left(\cos y\right)\cdot y'-3x^{2}y' & = & 6xy\\ \left(\cos y-3x^{2}\right)y' & = & 6xy\\ y' & = & \dfrac{6xy}{\cos y-3x^{2}} \end{array} \]