Implicit Differentiation

When an equation relating $x$ and $y$ cannot (or should not) be solved explicitly for $y$ , we can still find $d y / d x$ by differentiating both sides with respect to $x$ and treating $y$ as a differentiable function of $x$ .

Term	Meaning
Explicit function	$y = f (x)$ directly
Implicit function	$F (x, y) = 0$ , $y$ not isolated
Implicit differentiation	Differentiating $F (x, y) = 0$ w.r.t. $x$ , then solving for $d y / d x$

Explicit vs. Implicit Functions

In most examples so far, $y$ has been expressed explicitly in terms of $x$ :

y = \frac{\sin x}{x}, y = x \sqrt{x^{2} + 1}

Sometimes the relation has the form $F (x, y) = 0$ and $y$ is implicitly expressed:

x^{2} + y^{2} = 4, x^{3} - 3 x y + y^{3} = 0

The circle $x^{2} + y^{2} = 4$ implicitly defines two explicit functions:

f (x) = \sqrt{4 - x^{2}} (upper semicircle), g (x) = - \sqrt{4 - x^{2}} (lower semicircle)

Circle x^2+y^2=4 and its two semicircle functions

The curve $x^{3} - 3 x y + y^{3} = 0$ is the folium of Descartes.

It implicitly defines infinitely many functions and is difficult to solve for $y$ explicitly.

The Method of Implicit Differentiation

If we assume $y$ is a differentiable function of $x$ , we can apply $\frac{d}{d x}$ to both sides of $F (x, y) = 0$ , treating $y$ as a differentiable function of $x$ and using the Chain Rule wherever $y$ appears.

Key rule: $\frac{d}{d x} (y^{n}) = n y^{n - 1} \frac{d y}{d x}$ (Chain Rule with outer function $u^{n}$ and inner function $y = y (x)$ ).

Important: Only values $(x, y)$ satisfying the original relation can be substituted into the derivative.

Worked Examples

If $x^{2} + y^{2} = 4$ , find $d y / d x$ using explicit differentiation.

Solution

We can write

y = \sqrt{4 - x^{2}}

y = - \sqrt{4 - x^{2}}

. For

y = \sqrt{4 - x^{2}} = f (x)

: f'(x)=\frac{1}{2\sqrt{4-x^2}}(-2x)=-\frac{x}{\sqrt{4-x^2}}=-\frac{x}{y} For

y = - \sqrt{4 - x^{2}} = g (x)

: g'(x)=\frac{x}{\sqrt{4-x^2}}=-\frac{x}{g(x)}=-\frac{x}{y} In both cases,

\frac{d y}{d x} = - \frac{x}{y}

(when

y \neq 0

If $x^{2} + y^{2} = 4$ , find $d y / d x$ using implicit differentiation.

Solution

Assume

d y / d x

exists and apply

d / d x

to both sides:

\frac{d}{d x} (x^{2} + y^{2}) = \frac{d}{d x} 4

2x+2y\frac{dy}{dx}=0\tag{i} (For the

y^{2}

term we used the Chain Rule:

\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}

.) Solving equation (i) for

d y / d x

\frac{d y}{d x} = - \frac{x}{y}

Find the equation of the tangent line to $x^{3} + y^{3} - 3 x y = 0$ at $(\sqrt[3]{2}, \sqrt[3]{4})$ .

Solution

Differentiating implicitly:

3 x^{2} + 3 y^{2} \frac{d y}{d x} - 3 y - 3 x \frac{d y}{d x} = 0

\frac{d y}{d x} (3 y^{2} - 3 x) = 3 (y - x^{2})

\frac{d y}{d x} = \frac{y - x^{2}}{y^{2} - x}

(\sqrt[3]{2}, \sqrt[3]{4})

: y'=\frac{\sqrt[3]{4}-(\sqrt[3]{2})^2}{(\sqrt[3]{4})^2-\sqrt[3]{2}}=\frac{\sqrt[3]{4}-\sqrt[3]{4}}{\sqrt[3]{16}-\sqrt[3]{2}}=0 The tangent is horizontal:

y = \sqrt[3]{4}

Folium of Descartes with horizontal tangent

If $x^{2} - 3 x y + 2 y^{2} = 3$ , find $d y / d x$ .

Solution

Differentiating both sides:

2 x - 3 (y + x \frac{d y}{d x}) + 4 y \frac{d y}{d x} = 0

2 x - 3 y + (4 y - 3 x) \frac{d y}{d x} = 0

\frac{d y}{d x} = \frac{3 y - 2 x}{4 y - 3 x}

Higher-Order Derivatives via Implicit Differentiation

If $x^{2} + y^{2} = 4$ , find $d^{2} y / d x^{2}$ .

Solution

We already know

\frac{d y}{d x} = - \frac{x}{y}

. Applying the Quotient Rule: \begin{aligned} \frac{d^2y}{dx^2}&=\frac{d}{dx}\left(-\frac{x}{y}\right)=-\frac{1\cdot y-x\cdot\dfrac{dy}{dx}}{y^2}\\ &=-\frac{y-x\left(-\dfrac{x}{y}\right)}{y^2}=-\frac{y+\dfrac{x^2}{y}}{y^2}\\ &=-\frac{y^2+x^2}{y^3}=-\frac{4}{y^3} \end{aligned} (using

x^{2} + y^{2} = 4

in the last step)

If $y^{3} - x y - 1 = 0$ , find $d^{2} y / d x^{2}$ .

Solution

**Step 1:** Implicit differentiation for

d y / d x

3 y^{2} \frac{d y}{d x} - y - x \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{y}{3 y^{2} - x}

**Step 2:** Apply Quotient Rule:

\frac{d^{2} y}{d x^{2}} = \frac{\frac{d y}{d x} \cdot (3 y^{2} - x) - y \cdot (6 y \frac{d y}{d x} - 1)}{(3 y^{2} - x)^{2}}

Substituting

\frac{d y}{d x} = \frac{y}{3 y^{2} - x}

and simplifying:

\frac{d^{2} y}{d x^{2}} = \frac{- 2 x y}{(3 y^{2} - x)^{3}}

Given $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ (an ellipse), find $\frac{d^{2} y}{d x^{2}}$ .

Solution

**Step 1:** Implicit differentiation:

\frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{b^{2} x}{a^{2} y}

**Step 2:** Apply Quotient Rule:

\frac{d^{2} y}{d x^{2}} = - \frac{b^{2}}{a^{2}} \frac{y - x \cdot (- \frac{b^{2} x}{a^{2} y})}{y^{2}} = - \frac{b^{2}}{a^{4}} \frac{a^{2} y^{2} + b^{2} x^{2}}{y^{3}}

Since

b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2}

(from the ellipse equation multiplied by

a^{2} b^{2}

\frac{d^{2} y}{d x^{2}} = - \frac{b^{4}}{a^{2} y^{3}}

Frequently Asked Questions

When should I use implicit differentiation instead of solving for

y

first?

Use implicit differentiation when solving for

y

is difficult or impossible (e.g.,

x^{3} + y^{3} - 3 x y = 0

), or when it produces complicated expressions that would be harder to differentiate. Even when explicit formulas exist, implicit differentiation is often faster.

Can the implicit derivative formula fail?

Yes. The formula

d y / d x = - F_{x} / F_{y}

(partial derivatives) requires

F_{y} \neq 0

at the point. Geometrically this corresponds to the curve having a non-vertical tangent. At points where

F_{y} = 0

, the tangent may be vertical or the curve may have a corner.

Can any value of

(x, y)

be substituted into the implicit derivative?

Only values satisfying the original equation

F (x, y) = 0

. The derivative formula is derived under the assumption that

(x, y)

lies on the curve, so substituting off-curve values gives meaningless results.

Explicit vs. Implicit Functions

The Method of Implicit Differentiation

Worked Examples

Higher-Order Derivatives via Implicit Differentiation

Frequently Asked Questions

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