MISSION · Calculus
Implicit Differentiation
Differentiating a curve that was never written as y = something
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The idea in one sentence
Differentiate both sides of the equation as it stands, treating \(y\) as a function of \(x\), and then solve for \(\frac{dy}{dx}\).
The problem
\[x^2 + y^2 = 25\]This is a circle, and it plainly has a gradient at every point. But it isn’t a function: rearranging gives \(y = \pm\sqrt{25 - x^2}\), two separate half-circles, and you’d have to differentiate each and remember which one you’re on. Worse cases like \(x^3 + y^3 = 6xy\) can’t be rearranged at all.
Implicit differentiation avoids the rearranging entirely.
The method
Differentiate every term with respect to \(x\). The only new idea is what to do with \(y\) terms, and it’s the chain rule:
\[\frac{d}{dx}\left(y^2\right) = 2y \cdot \frac{dy}{dx}\]Because \(y\) is a function of \(x\), differentiating anything built from \(y\) produces a factor of \(\frac{dy}{dx}\). That factor is the whole technique.
\[2x + 2y\frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{x}{y}\]At \((3, 4)\) the gradient is \(-3/4\). Sanity check: the radius to that point has gradient \(4/3\), and the tangent should be perpendicular to it. Perpendicular gradients multiply to \(-1\), and \(\frac{4}{3} \times -\frac{3}{4} = -1\). The formula is right.
Reading the answer
\(\frac{dy}{dx} = -\frac{x}{y}\) depends on both coordinates, which is normal for implicit results and is exactly what you want: a circle has two points above each \(x\), with opposite gradients, and the \(y\) in the answer is what distinguishes them.
At \(y = 0\) the gradient is undefined — the far left and right of the circle, where the tangent is vertical. The formula tells you that without being asked.
Products need the product rule
\[x^3 + y^3 = 6xy\]The right side is a product of two functions of \(x\):
\[3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}\]Gather the \(\frac{dy}{dx}\) terms on one side, factor, divide:
\[\frac{dy}{dx}\left(3y^2 - 6x\right) = 6y - 3x^2 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}\]That collect-and-factor step is the same every time, whatever the equation.
Where else it turns up
Differentiating \(y = \ln x\) is easiest done implicitly: write \(e^y = x\), differentiate to get \(e^y \frac{dy}{dx} = 1\), so \(\frac{dy}{dx} = 1/e^y = 1/x\). The same trick gives the derivatives of the inverse trigonometric functions.
And related rates is implicit differentiation with respect to time rather than \(x\) — same chain rule, different variable.
Common mix-ups
- Forgetting the \(\frac{dy}{dx}\) factor on \(y\) terms, which is the error that makes the whole thing collapse.
- Not using the product rule on mixed terms like \(xy\), whose derivative is \(y + x\frac{dy}{dx}\).
- Leaving \(\frac{dy}{dx}\) on both sides instead of collecting it.
- Expecting an answer in \(x\) alone. Most implicit derivatives keep both variables.
Check yourself
Differentiate \(x^2 + y^2 = 25\) implicitly, then find the gradient at \((3, 4)\).
\(2x + 2y\frac{dy}{dx} = 0\) gives \(\frac{dy}{dx} = -x/y\). At (3, 4) that is −3/4, which is perpendicular to the radius of gradient 4/3 — as a tangent to a circle must be.
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