Implicit Differentiation

Implicit differentiation is simply a way to find the slope of a curve when its equation isn't written as "$y =$ something." ("Implicit" because the equation only implies how $y$ depends on $x$ instead of saying it outright.) It's useful for circles, ellipses and all sorts of constraints where solving for $y$ is messy or impossible. The formula $\frac{d}{dx}[y^n] = ny^{n-1}y'$ is just the power rule, $ny^{n-1}$ (bring the exponent down as usual), times $y'$ (the chain rule's inner derivative, since $y$ is a function of $x$). That's really the only new idea, so bear with it.

Some curves, like the circle $x^2 + y^2 = 25$, aren't the graph of a single function. Each $x$ between $-5$ and $5$ has two $y$ values, one on top and one on the bottom. You could split it into $y = \pm\sqrt{25 - x^2}$ and differentiate each half, but that gets ugly fast, and for curves like $x^3 + y^3 = 6xy$ there's no nice way to solve for $y$ at all. Still, the curves clearly have tangent lines. So what can we do?

Here's the idea. Near any point on the curve, $y$ really is some function of $x$, even if we can't write it down. So just differentiate the equation as it stands, and treat $y$ the way the chain rule says to treat an inside function. Differentiating $y^2$ gives $2y\cdot y'$, not just $2y$, for the same reason $\sin(5x)$ gives $5\cos(5x)$.

Steps:

1. Differentiate both sides with respect to $x$. Each time you differentiate an expression in $y$, multiply by $y'$.
2. Collect the $y'$ terms on one side.
3. Factor out $y'$ and divide.

Products like $xy$ need the product rule: $\frac{d}{dx}(xy) = y + xy'$.

The answer usually contains both $x$ and $y$. That's expected, since the slope depends on where you are on the curve, and one $x$ can correspond to several points. To get a number, plug in a specific point on the curve.

Physics constraints are equations, not functions: a pendulum on a fixed-length rod, a planet on an ellipse. Implicit differentiation gives velocities and slopes directly from the constraint.