Implicit differentiation
Master implicit differentiation for Year 12 Specialist Mathematics in Queensland (QCAA). When a curve such as \(x^2+y^2=25\) or \(xy=12\) is not solved for \(y\), you differentiate both sides with respect to \(x\) and use the chain rule on every \(y\)-term to find its gradient.
You will learn to differentiate circles, ellipses and \(xy=k\) relations, collect and factor the \(\dfrac{dy}{dx}\) terms, and find gradients, tangents and normals to implicit curves — a core technique of Unit 4 rates of change and the foundation for related rates.
Theory
Implicit differentiation finds \(\dfrac{dy}{dx}\) for a curve given by a relation such as \(x^2+y^2=25\) that is not solved for \(y\). In Year 12 Specialist Mathematics (QCAA, Queensland) you differentiate both sides with respect to \(x\), using the chain rule on every \(y\)-term, then make \(\dfrac{dy}{dx}\) the subject. This page shows the method and uses it to find gradients, tangents and normals to circles, ellipses and \(xy=k\) curves.
A relation is written in implicit form when \(x\) and \(y\) are mixed together and \(y\) is not made the subject — for example the circle \(x^2+y^2=25\), the ellipse \(4x^2+9y^2=36\), or the rectangular hyperbola \(xy=12\). Such a curve still defines \(y\) as a function of \(x\) locally, so it has a gradient \(\dfrac{dy}{dx}\) at each point.
Implicit differentiation differentiates both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\). Every term in \(x\) differentiates normally; every term in \(y\) needs the chain rule, so \(\dfrac{d}{dx}\big(y^n\big)=n\,y^{\,n-1}\dfrac{dy}{dx}\).
A product such as \(xy\) needs the product rule: \(\dfrac{d}{dx}(xy)=x\dfrac{dy}{dx}+y\). After differentiating, collect the \(\dfrac{dy}{dx}\) terms on one side and factor, then divide to make \(\dfrac{dy}{dx}\) the subject. The answer is usually in terms of both \(x\) and \(y\).
The gradient at a point is found by substituting the point's coordinates into \(\dfrac{dy}{dx}\); from there the tangent uses \(y-y_1=m(x-x_1)\) and the normal uses the negative reciprocal gradient \(-\dfrac{1}{m}\).
Differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\). The key chain-rule result for a power of \(y\) is:
In particular \(\dfrac{d}{dx}(y^2)=2y\dfrac{dy}{dx}\) and \(\dfrac{d}{dx}(y)=\dfrac{dy}{dx}\). A product of \(x\) and \(y\) uses the product rule:
For a circle \(x^2+y^2=r^2\) the method gives the standard gradient:
How to differentiate implicitly
- Differentiate both sides with respect to \(x\), term by term. Treat \(y\) as a function of \(x\).
- Apply the chain rule to every \(y\)-term, so \(y^2\) becomes \(2y\dfrac{dy}{dx}\); use the product rule for any \(xy\) term.
- Collect and factor: gather all \(\dfrac{dy}{dx}\) terms on one side, factor out \(\dfrac{dy}{dx}\), then divide to make it the subject.
- Substitute if asked: put the point's coordinates in for the gradient, then use \(y-y_1=m(x-x_1)\) for a tangent, or \(-\dfrac{1}{m}\) for a normal.
Differentiate each term with respect to \(x\); by the chain rule \(\dfrac{d}{dx}(y^2)=2y\dfrac{dy}{dx}\):
| \(\dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(y^2)\) | \(=\) | \(\dfrac{d}{dx}(9)\) |
| \(2x+2y\dfrac{dy}{dx}\) | \(=\) | \(0\) |
| \(2y\dfrac{dy}{dx}\) | \(=\) | \(-2x\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(-\dfrac{2x}{2y}\) |
| \(=\) | \(-\dfrac{x}{y}\) |
\(\dfrac{dy}{dx}=-\dfrac{x}{y}\).
The term \(xy\) needs the product rule \(\dfrac{d}{dx}(xy)=x\dfrac{dy}{dx}+y\), and \(\dfrac{d}{dx}(10)=0\):
| \(\dfrac{d}{dx}(xy)\) | \(=\) | \(\dfrac{d}{dx}(10)\) |
| \(x\dfrac{dy}{dx}+y\) | \(=\) | \(0\) |
| \(x\dfrac{dy}{dx}\) | \(=\) | \(-y\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(-\dfrac{y}{x}\) |
\(\dfrac{dy}{dx}=-\dfrac{y}{x}\).
Differentiate implicitly, using \(\dfrac{d}{dx}(4y^2)=8y\dfrac{dy}{dx}\), and make \(\dfrac{dy}{dx}\) the subject:
| \(2x+8y\dfrac{dy}{dx}\) | \(=\) | \(0\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(-\dfrac{2x}{8y}\) |
| \(=\) | \(-\dfrac{x}{4y}\) |
Substitute the point \((3,1)\):
| \(\dfrac{dy}{dx}\) | \(=\) | \(-\dfrac{3}{4(1)}\) |
| \(=\) | \(-\dfrac{3}{4}\) |
The gradient at \((3,1)\) is \(-\dfrac{3}{4}\).
Differentiate implicitly and substitute \((2,4)\) for the gradient:
| \(2x+2y\dfrac{dy}{dx}\) | \(=\) | \(0\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(-\dfrac{x}{y}\) |
| \(m\) | \(=\) | \(-\dfrac{2}{4}\) |
| \(=\) | \(-\dfrac{1}{2}\) |
Use point-gradient form \(y-y_1=m(x-x_1)\) and clear fractions:
| \(y-4\) | \(=\) | \(-\dfrac{1}{2}(x-2)\) |
| \(2y-8\) | \(=\) | \(-(x-2)\) |
| \(2y-8\) | \(=\) | \(-x+2\) |
| \(x+2y\) | \(=\) | \(10\) |
The tangent is \(x+2y=10\).
Common pitfalls
Frequently asked questions
What is implicit differentiation?
It is a way to find \(\dfrac{dy}{dx}\) for a curve whose equation mixes \(x\) and \(y\) and is not solved for \(y\). You differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\).
Why does \(y^2\) differentiate to \(2y\dfrac{dy}{dx}\)?
Because \(y\) is a function of \(x\), the chain rule applies: \(\dfrac{d}{dx}(y^2)=2y\times\dfrac{dy}{dx}\). The extra factor \(\dfrac{dy}{dx}\) is what makes implicit differentiation different from ordinary differentiation.
How do I differentiate \(xy\)?
Use the product rule: \(\dfrac{d}{dx}(xy)=x\dfrac{dy}{dx}+y\). One factor is \(x\) and the other is \(y\), which is itself a function of \(x\).
How do I find the gradient of a circle at a point?
Differentiate \(x^2+y^2=r^2\) implicitly to get \(\dfrac{dy}{dx}=-\dfrac{x}{y}\), then substitute the coordinates of the point. For example at \((3,4)\) on \(x^2+y^2=25\) the gradient is \(-\dfrac{3}{4}\).
How do I get the tangent from the gradient?
Find the gradient \(m\) at the point, then use point-gradient form \(y-y_1=m(x-x_1)\) and rearrange. For the normal, use the negative reciprocal gradient \(-\dfrac{1}{m}\).
Do I need to solve the equation for \(y\) first?
No. That is the whole point of implicit differentiation — you differentiate the relation as it is written, which is much easier than solving for \(y\) and often the only practical option.