Calculus Differentiation

Implicit differentiation

Q: What is implicit differentiation?

A way to find dy dx for an equation in x and y without solving for y. Differentiate both sides, attach a dy dx to every y term, then solve.

Q: Why does a dy/dx appear when differentiating y-terms?

Because y is a function of x, so the chain rule applies: the derivative of y squared is 2y times dy dx.

Q: How do you find the tangent to a circle?

Differentiate x squared plus y squared equals r squared implicitly to get slope minus x over y, plug in the point of tangency, then use point slope form.

Q: When do I need implicit differentiation instead of the normal rules?

When the equation mixes x and y so that solving for y is awkward or impossible: circles, ellipses, and many physics or economics relations.

20 practice questions 0 video lessons Theory + worked examples

Implicit Differentiation

California Calculus • Standard 6.0 • Differentiation

Implicit Differentiation is a topic in Differentiation in the California Calculus Standards. It is aligned to Standard 6.0, which requires students to use implicit differentiation in a wide variety of problems.

Implicit differentiation finds \(\dfrac{dy}{dx}\) for a curve given by an equation in \(x\) and \(y\), such as a circle, by differentiating both sides with the chain rule and solving for \(\dfrac{dy}{dx}\).

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Theory

Implicit differentiation finds \(\dfrac{dy}{dx}\) for a curve given by an equation in \(x\) and \(y\) — like a circle — without first solving for \(y\). Differentiate both sides, remembering that every \(y\) carries a hidden \(\dfrac{dy}{dx}\) factor from the chain rule, then solve.

Some curves cannot easily be written as \(y=f(x)\); the circle \(x^2+y^2=25\) is one. Implicit differentiation differentiates the equation as it stands, treating \(y\) as a function of \(x\).

The key move is the chain rule: differentiating a \(y\)-term produces a \(\dfrac{dy}{dx}\) factor. For example \(\dfrac{d}{dx}\big(y^2\big)=2y\,\dfrac{dy}{dx}\).

After differentiating, collect the \(\dfrac{dy}{dx}\) terms and solve for \(\dfrac{dy}{dx}\).

Key idea: the slope usually depends on both \(x\) and \(y\). To get a number, substitute a point that lies on the curve.

Tangent to the circle \(x^2+y^2=25\); slope \(=-\dfrac{x}{y}\).

The same method gives the tangent to an ellipse.

The building block — differentiating a power of \(y\):

\[\dfrac{d}{dx}\big(y^{n}\big)=n\,y^{\,n-1}\,\dfrac{dy}{dx}\]

derivative of y to the n is n y to the n minus one times dy dx

For the circle \(x^2+y^2=r^2\), differentiating gives

\[2x+2y\,\dfrac{dy}{dx}=0\ \Rightarrow\ \dfrac{dy}{dx}=-\dfrac{x}{y}\]

circle slope is minus x over y

Product and quotient terms still need their rules: \(\dfrac{d}{dx}(xy)=y+x\,\dfrac{dy}{dx}\).

How to differentiate implicitly

Differentiate both sides with respect to \(x\), attaching a \(\dfrac{dy}{dx}\) to every \(y\)-term (chain rule).
Collect all \(\dfrac{dy}{dx}\) terms on one side.
Solve for \(\dfrac{dy}{dx}\); substitute a point if you need a numeric slope.

Example 1 — The unit-style circle

Find \(\dfrac{dy}{dx}\) for \(x^2+y^2=25\).

Solution

Differentiate both sides, treating \(y\) as \(y(x)\) (so \(y^2\) gives \(2y\,y'\)).

\(2x+2y\,y'\)	\(=\)	\(0\)
\(y'\)	\(=\)	\(-\dfrac{x}{y}\)

dy dx equals minus x over y

Example 2 — Slope at a point

Find the slope of \(x^2+y^2=25\) at \((3,4)\).

Solution

Use \(y'=-\dfrac{x}{y}\) from Example 1.

\(y'\)

\(=\)

\(-\dfrac{3}{4}\)

The tangent at \((3,4)\) has slope \(-\dfrac{3}{4}\).

slope is negative three quarters

Example 3 — A product term

Find \(\dfrac{dy}{dx}\) for \(xy=6\).

Solution

Differentiate \(xy\) with the product rule: \(y+x\,y'\).

\(y+x\,y'\)	\(=\)	\(0\)
\(y'\)	\(=\)	\(-\dfrac{y}{x}\)

dy dx equals minus y over x

Example 4 — An ellipse

Find \(\dfrac{dy}{dx}\) for \(x^2+4y^2=8\).

Solution

Differentiate each term; \(4y^2\) gives \(8y\,y'\).

\(2x+8y\,y'\)	\(=\)	\(0\)
\(y'\)	\(=\)	\(-\dfrac{x}{4y}\)

dy dx equals minus x over four y

Common pitfalls

Every \(y\) gets a \(\dfrac{dy}{dx}\). Differentiating \(y^2\) is \(2y\,\dfrac{dy}{dx}\), not just \(2y\).

Use the product rule on mixed terms. \(\dfrac{d}{dx}(xy)=y+x\,y'\), not \(y'\).

The slope needs a point on the curve. \(-\dfrac{x}{y}\) is only a number once you plug in coordinates that satisfy the equation.

Frequently asked questions

What is implicit differentiation?

A way to find \(\dfrac{dy}{dx}\) for an equation in \(x\) and \(y\) without solving for \(y\). Differentiate both sides, attach a \(\dfrac{dy}{dx}\) to every \(y\)-term, then solve.

Why does a dy/dx appear when differentiating y-terms?

Because \(y\) is a function of \(x\), so the chain rule applies: \(\dfrac{d}{dx}(y^2)=2y\cdot\dfrac{dy}{dx}\).

How do you find the tangent to a circle?

Differentiate \(x^2+y^2=r^2\) implicitly to get slope \(-\dfrac{x}{y}\), plug in the point of tangency, then use point-slope form.

When do I need implicit differentiation instead of the normal rules?

When the equation mixes \(x\) and \(y\) so that solving for \(y\) is awkward or impossible — circles, ellipses, and many physics/economics relations.

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