Implicit Differentiation
Roll a ball around the inside of a circular bowl and, at every instant, it is heading in some
direction — it has a slope. Yet the circle it traces,
x^2 + y^2 = 25, is not a function: above most values of
x there sit two heights, one on the top arc and one on the
bottom. You cannot write it as a single tidy y = f(x). So how do you
differentiate a curve that refuses to be solved for y?
The answer is a beautifully sneaky move called implicit differentiation: you
differentiate the equation just as it stands, treating y as a
hidden function of x, and let the chain rule do the untangling. It
unlocks the slope of circles, ellipses, and every knotted curve that a plain
y=f(x) could never describe.
So far every curve has come solved for y:
y = x^2, y = (3x+1)^5. But some curves
tangle y up with x and refuse to be
untangled. The circle of radius 5 is the classic example:
x^2 + y^2 = 25
We can still find the slope \frac{dy}{dx} — without ever solving
for y. The trick is to differentiate both sides with
respect to x, treating y as a
(hidden) function of x. Then every y-term
picks up an extra \frac{dy}{dx} by the
chain rule:
\frac{d}{dx}\bigl[y^2\bigr] = 2y\cdot\frac{dy}{dx}
That is the chain rule with an outer function u^2 and an inner
function y: bring the power down (2y),
then multiply by the derivative of the inside (\frac{dy}{dx}).
Differentiating the whole equation term by term:
2x + 2y\,\frac{dy}{dx} = 0
Now just solve for \frac{dy}{dx} — subtract
2x and divide by 2y:
\frac{dy}{dx} = -\frac{x}{y}
One formula gives the slope at every point of the circle. At
(3,4) the slope is
-\tfrac{3}{4}; at (4,3) it is
-\tfrac{4}{3}. No solving for y required.
The recipe
To differentiate an equation that mixes x and
y together:
-
Differentiate every term on both sides with respect to
x.
-
For a term involving y, differentiate as usual and then
multiply by \frac{dy}{dx} — that extra factor
is the chain rule, since y is itself a function of
x.
-
Collect all the \frac{dy}{dx} terms on one
side, factor it out, and solve for
\frac{dy}{dx}.
This is exactly how you find the slope of a curve — a circle, an ellipse, any relation — that
isn't a function and so can never be written as y = f(x).
A mixed term: the product rule strikes
Not every equation splits into neat separate x- and
y-terms. Consider
x^2 + xy + y^2 = 7.
The middle term xy is a product of two things that both
depend on x, so it needs the
product rule. Watch it
differentiate term by term:
\underbrace{2x}_{x^2} \;+\; \underbrace{\Bigl(1\cdot y + x\cdot\tfrac{dy}{dx}\Bigr)}_{xy\ \text{(product rule)}} \;+\; \underbrace{2y\,\tfrac{dy}{dx}}_{y^2} \;=\; 0.
Gather the \frac{dy}{dx} terms on one side and everything else on the
other:
x\,\frac{dy}{dx} + 2y\,\frac{dy}{dx} = -2x - y.
Factor out \frac{dy}{dx} and divide:
\frac{dy}{dx} = \frac{-2x - y}{x + 2y}.
Notice the pattern that appears again and again: differentiate, collect the
\frac{dy}{dx} terms, factor, divide. That closing manoeuvre is
the signature of every implicit-differentiation problem.
Worked example: a tangent line
Find the equation of the tangent to x^2 + y^2 = 25 at the point
(-3, 4).
Step 1 — the slope. We already have
\frac{dy}{dx} = -\frac{x}{y}. Substitute the point:
\left.\frac{dy}{dx}\right|_{(-3,4)} = -\frac{-3}{4} = \frac{3}{4}.
Step 2 — the line. Using point–slope form
y - y_1 = m(x - x_1):
y - 4 = \tfrac{3}{4}\,(x + 3).
A quick sanity check: the radius to (-3,4) has slope
\tfrac{4}{-3} = -\tfrac{4}{3}, and the tangent's slope
\tfrac{3}{4} is its negative reciprocal — so the tangent is
perpendicular to the radius, exactly as a circle demands. The geometry confirms the calculus.
The slope of a circle, pictured
Here is x^2 + y^2 = 25 with the tangent line at
(3,4). Implicit differentiation predicts a slope of
-\frac{x}{y} = -\frac{3}{4} there — and the tangent line bears it
out. Notice the tangent is perpendicular to the radius, just as it should be.
Three slips account for nearly every lost mark on implicit differentiation.
-
Never forget the \frac{dy}{dx} on a y-term.
Differentiating y^3 gives
3y^2\cdot\frac{dy}{dx}, not just
3y^2. That missing factor is the chain rule — y
is a function of x — and dropping it is the classic error.
-
A term like xy needs the product rule, giving
y + x\frac{dy}{dx}. It is not
\frac{dy}{dx}, and it is not just y.
-
Collect before you divide. After differentiating you'll have
\frac{dy}{dx} scattered across the equation. Gather every one of those
terms on a single side, factor it out, and only then divide. Trying to solve before
collecting turns easy algebra into a mess.
Implicit differentiation is the secret engine behind related rates — problems
where several changing quantities are chained together by one equation. A ladder leans against a
wall; its foot slides out at a known speed. How fast does its top slide down? Pythagoras ties the
two together, x^2 + y^2 = L^2, and differentiating implicitly
with respect to time gives
2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 — instantly relating the two
speeds. The same trick tells you how fast a balloon's radius grows as you pump air in, or how the
shadow of a walking person lengthens.
It reaches further still. Differentiating through webs of interdependent variables is exactly what
backpropagation does inside a neural network — chaining derivatives through
thousands of tangled quantities to work out which knob to turn. When you meet
related rates,
you'll recognise this very move dressed in new clothes.
See it explained