Implicit differentiation
Master implicit differentiation in Year 12 VCE Specialist Mathematics. 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. It sits in the Calculus area of study of the VCE Mathematics Study Design (VCAA), within the Differential calculus topic of Unit 3.
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 that also underpins 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 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.