When you need it
Some curves cannot be written as \(y = f(x)\). The circle \(x^{2} + y^{2} = 25\) fails the vertical line test, so no single function describes it. Implicit differentiation finds \(dy/dx\) anyway, without ever solving for \(y\).
The idea: treat \(y\) as an unknown function of \(x\), differentiate both sides of the equation, and use the chain rule every time a \(y\) appears — because differentiating \(y\) with respect to \(x\) gives \(dy/dx\), not 1.
The method
Move everything to one side so the equation reads \(F(x, y) = 0\). Then:
- Differentiate both sides with respect to \(x\), attaching \(\frac{dy}{dx}\) to every term that came from a \(y\).
- Collect all the \(\frac{dy}{dx}\) terms on one side, everything else on the other.
- Factor out \(\frac{dy}{dx}\) and divide.
The shortcut this calculator uses is the same result in one line:
\[\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y}\]
It follows from the implicit function theorem, and it is why the two partial derivatives are shown alongside the answer. Wherever \(\partial F/\partial y = 0\), the tangent is vertical and \(dy/dx\) does not exist.
Worked example: the circle
Take the circle \(x^{2} + y^{2} = 25\), so \(F = x^{2} + y^{2} - 25\).
Differentiating term by term with respect to \(x\):
\[2x + 2y\frac{dy}{dx} = 0\]
The \(2y\frac{dy}{dx}\) comes from the chain rule — \(y^{2}\) differentiates to \(2y\) times the derivative of the inside. Rearranging:
\[\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}\]
Sanity check: at \((3, 4)\) the slope is \(-3/4\), perpendicular to the radius, exactly as a circle requires. At \((5, 0)\) the denominator vanishes — and that is the point where the tangent is vertical.
Worked example: the folium of Descartes
The folium of Descartes, \(x^{3} + y^{3} = 6xy\), is the standard harder example. Writing \(F = x^{3} + y^{3} - 6xy\) and differentiating gives:
\[3x^{2} + 3y^{2}\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}\]
Collecting and factoring:
\[\frac{dy}{dx} = \frac{6y - 3x^{2}}{3y^{2} - 6x} = \frac{2y - x^{2}}{y^{2} - 2x}\]
Notice the answer contains both \(x\) and \(y\). That is normal for implicit differentiation — to get a number you need a point on the curve, not just an \(x\).
Things to watch
- Enter the equation with everything moved to one side. \(x^{2}+y^{2}=25\) is typed as x^2 + y^2 - 25.
- The answer normally contains both variables. Substitute a point on the curve to get a slope.
- Where \(\partial F/\partial y = 0\) the formula divides by zero — that is a vertical tangent, not an error in your working.
- Implicit differentiation is also how the inverse function derivatives are proved: start from \(\sin y = x\) and differentiate both sides.
Questions
What is implicit differentiation?
A way of finding \(dy/dx\) for a curve given by an equation in \(x\) and \(y\) that has not been solved for \(y\), such as \(x^{2} + y^{2} = 25\).
How do I enter an equation?
Move everything to one side and type that expression. For \(x^{2}+y^{2}=25\), enter x^2 + y^2 - 25.
Why does the answer still contain y?
Because the curve has different slopes at different points with the same \(x\). Substituting a point \((x, y)\) that lies on the curve gives the slope there.
When does implicit differentiation fail?
Where \(\partial F/\partial y = 0\). The formula divides by zero, which corresponds to a vertical tangent — the slope genuinely does not exist there.
Is this on AP Calculus?
Yes, implicit differentiation is on both AP Calculus AB and BC, and it is a standard Calculus 1 topic.