Implicit differentiation

A relation such as \(x^2+xy+y^2=7\) describes a curve even though it is not solved for \(y\), and we can still find its slope. Implicit differentiation works because \(y\) is treated as a function of \(x\), and when a term involving \(y\) is differentiated with respect to \(x\), the chain rule contributes a factor of \(dy/dx\).

The chain rule is the method

Differentiate \(x^2+xy+y^2=7\) with respect to \(x\). The first term gives \(2x\), the product \(xy\) requires the product rule, \(\tfrac{d}{dx}(xy)=y+x\tfrac{dy}{dx}\), and the last term requires the chain rule, \(\tfrac{d}{dx}(y^2)=2y\tfrac{dy}{dx}\).

So \(2x+y+x\tfrac{dy}{dx}+2y\tfrac{dy}{dx}=0\), and collecting the derivative terms gives

\[\begin{aligned} \left(x+2y\right)\frac{dy}{dx} &= -(2x+y)\\ \frac{dy}{dx} &= -\frac{2x+y}{x+2y} \end{aligned}\]

The extra \(dy/dx\) factors are not a special rule for implicit differentiation. They are ordinary chain-rule factors.

The slope depends on both coordinates

An explicit derivative such as \(f'(x)=2x\) depends only on \(x\), while an implicit derivative can depend on both \(x\) and \(y\). That matters because a vertical line may meet an implicit curve at more than one point.

For the relation above, setting \(x=1\) gives \(y^2+y-6=0\), so \(y=2\) or \(y=-3\). At \((1,2)\), \(\tfrac{dy}{dx} = -\tfrac{2+2}{1+4} = -\tfrac45\), and at \((1,-3)\), \(\tfrac{dy}{dx} = -\tfrac{2-3}{1-6} = -\tfrac15\). The two points have the same \(x\)-coordinate and different tangent slopes, so when an implicit problem asks for a slope, the full point usually matters.

The visualization shows several implicit relations without solving them for \(y\). The point moves along the curve, and the tangent line is drawn from the slope predicted by the implicit derivative. The reported slope can also be checked by parametrizing the curve and computing \(\tfrac{dy}{dx} = \tfrac{dy/dt}{dx/dt}\), and the two methods agree wherever the slope exists.

When the denominator is zero

For the circle \(x^2+y^2=25\), implicit differentiation gives \(\tfrac{dy}{dx}=-\tfrac{x}{y}\). At \((5,0)\), the denominator is zero. That does not mean the curve itself is undefined there. It means the tangent is vertical, so its slope is not a finite number, and the algebra is reporting the geometry correctly.

This is also why critical points of an implicit relation include points where the derivative is zero and points where the derivative does not exist. A zero numerator may produce a horizontal tangent, and a zero denominator may produce a vertical tangent.

Second derivatives stay implicit

Differentiating again often leaves \(x\), \(y\), and \(dy/dx\) in the expression. For the circle, \(\tfrac{dy}{dx}=-\tfrac{x}{y}\), and differentiating using the quotient rule gives

\[\frac{d^2y}{dx^2} = -\frac{y-x\frac{dy}{dx}}{y^2}\]

Now substitute \(\tfrac{dy}{dx}=-\tfrac{x}{y}\), so

\[\frac{d^2y}{dx^2} = -\frac{y+\frac{x^2}{y}}{y^2} = -\frac{x^2+y^2}{y^3}\]

Since the original relation gives \(x^2+y^2=25\), we obtain \(\tfrac{d^2y}{dx^2} = -\tfrac{25}{y^3}\). The original relation is often useful after the differentiation because it can simplify the result.