Implicit Differentiation (College Board AP® Calculus BC): Study Guide

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on 4 June 2025

Derivatives of implicit functions

What is an implicit function?

An equation in the form $y = f (x)$ or $x = f (y)$ is said to be written explicitly
- E.g. $y = 3 x^{2} + 2 x - 3$
Equations in terms of both $x$ and $y$ which cannot be written as $y$ in terms of $x$ are referred to as implicit functions
- E.g. $3 x^{2} - 7 x y^{2} = 3$ or $x^{2} + y^{2} = 25$
- For such an equation
  - We cannot write the relationship between $x$ and $y$ explicitly by expressing $y$ as a function of $x$
  - But values of $y$ satisfying the equation still depend implicitly on the values of $x$

What is implicit differentiation?

The method by which implicit functions are differentiated is known as implicit differentiation
To differentiate an implicit function with respect to $x$ , each term is differentiated with respect to $x$
For terms only involving $x$ , this is straightforward
However, to differentiate an expression that is in terms of $y$ , but differentiate it with respect to $x$ , we apply the chain rule
- $\frac{d}{d x} f (y) = f^{'} (y) \cdot y^{'} = f^{'} (y) \cdot \frac{d y}{d x}$
- In short, this means:
  - Differentiate the function that is in terms of $y$ , with respect to $y$ ,
  - and then multiply it by the term $\frac{d y}{d x}$
This could also be written as
- $\frac{d u}{d x} = \frac{d u}{d y} \cdot \frac{d y}{d x}$ where $u$ is a function in terms of $y$
Once each term has been differentiated with respect to $x$ ,
- rearrange to make $\frac{d y}{d x}$ the subject
- you may have to factorize out $\frac{d y}{d x}$ to do this
The final expression for $\frac{d y}{d x}$ will often be in terms of both $x$ and $y$
- To then find the value of the derivative at a point, you would need the coordinates of a point $(x, y)$ to substitute in
  - And substitute in both the $x$ and $y$ values
- This is only valid if the point $(x, y)$ lies on the curve in question
  - To check this, substitute the $x$ and $y$ values into the original equation of the curve, and check that the equation is satisfied
    - If it is then the point lies on the curve
    - If it is not, then the point does not lie on the curve

How do I use implicit differentiation?

For example, to differentiate $x^{2} + y^{2} = 4 x$
The derivative would be $\frac{d}{d x} (x^{2}) + \frac{d}{d x} (y^{2}) = \frac{d}{d x} (4 x)$
The terms which are only in terms of $x$ are straight forward
- $2 x + \frac{d}{d x} (y^{2}) = 4$
Then apply the result $\frac{d}{d x} f (y) = f^{'} (y) \cdot \frac{d y}{d x}$ to the remaining term, which is in terms of $y$
- $2 x + (2 y \cdot \frac{d y}{d x}) = 4$
Rearrange to find an expression for $\frac{d y}{d x}$
- $\frac{d y}{d x} = \frac{4 - 2 x}{2 y} = \frac{2 - x}{y}$

Worked Example

A curve is defined by the equation

$4 x^{3} + 2 x + 2 y^{3} + 6 y = 44$

(a) Confirm that the point $(2, 1)$ lies on the curve.

Answer:

Substitute $x = 2$ and $y = 1$ into the equation

$\begin{array}{rcl} 4 {(2)}^{3} + 2 (2) + 2 {(1)}^{3} + 6 (1) & = & 44 \\ 32 + 4 + 2 + 6 & = & 44 \\ 44 & = & 44 \end{array}$

The equation is satisfied, so the point lies on the curve

$(2, 1)$ lies on the curve

(b) Find the value of the derivative $\frac{d y}{d x}$ at the point $(2, 1)$ .

Answer:

Differentiate each term individually with respect to $x$

Use the result $\frac{d}{d x} f (y) = f^{'} (y) \cdot \frac{d y}{d x}$ for the $y$ -terms

Remember that the constant 44 on the right-hand side of the equation will differentiate to zero

$12 x^{2} + 2 + 6 y^{2} \cdot \frac{d y}{d x} + 6 \cdot \frac{d y}{d x} = 0$

Factorize out the $\frac{d y}{d x}$ term

$12 x^{2} + 2 + \frac{d y}{d x} (6 y^{2} + 6) = 0$

Rearrange to make $\frac{d y}{d x}$ the subject and simplify

$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{- 12 x^{2} - 2}{6 y^{2} + 6} \\ \frac{d y}{d x} & = & \frac{- 6 x^{2} - 1}{3 y^{2} + 3} \end{array}$

To find the value of the derivative at $(2, 1)$ substitute in $x = 2$ and $y = 1$

$\frac{d y}{d x} = \frac{- 6 (2^{2}) - 1}{3 (1^{2}) + 3} = \frac{- 24 - 1}{6} = - \frac{25}{6}$

The value of the derivative at $(2, 1)$ is $- \frac{25}{6}$

How might implicit differentiation questions be made harder?

Implicit differentiation may be combined with other skills including
- Chain rule
- Product rule
- Quotient rule
- Derivatives of exponentials, logarithms, and trigonometric functions
The following result can be useful when using the product rule
- $\frac{d}{d x} (x y) = y + x \frac{d y}{d x}$

Worked Example

Given that $x^{4} + y^{3} = 12 x y$ , find an expression for $\frac{d y}{d x}$ .

Answer:

Differentiate each term individually with respect to $x$

Use the result $\frac{d}{d x} f (y) = f^{'} (y) \cdot \frac{d y}{d x}$ for the terms with $y$

$4 x^{3} + 3 y^{2} \cdot \frac{d y}{d x} = \frac{d}{d x} (12 x y)$

To differentiate $12 x y$ with respect to $x$ , use the product rule

$u = 12 x$ and $v = y$

$u^{'} = 12$ and $v^{'} = 1 \cdot \frac{d y}{d x}$

$u^{'} v + u v^{'} = 12 y + 12 x \cdot \frac{d y}{d x}$

So the differentiated expression is

$4 x^{3} + 3 y^{2} \cdot \frac{d y}{d x} = 12 y + 12 x \cdot \frac{d y}{d x}$

Rearrange to make $\frac{d y}{d x}$ the subject

$\begin{array}{rcl} 3 y^{2} \cdot \frac{d y}{d x} - 12 x \cdot \frac{d y}{d x} & = & 12 y - 4 x^{3} \\ \frac{d y}{d x} (3 y^{2} - 12 x) & = & 12 y - 4 x^{3} \\ \frac{d y}{d x} & = & \frac{12 y - 4 x^{3}}{3 y^{2} - 12 x} \end{array}$

$\frac{d y}{d x} = \frac{12 y - 4 x^{3}}{3 y^{2} - 12 x}$

Equivalent answers would also be correct, for example:

$\frac{d y}{d x} = \frac{4 x^{3} - 12 y}{12 x - 3 y^{2}}$

Worked Example

Find $\frac{d y}{d x}$ given that $\cos (x + y) = 2 x^{2}$ .

Answer:

Differentiate both sides with respect to $x$

Use the chain rule for the term $\cos (x + y)$

The chain rule states that $f (g (x))$ differentiates to $f^{'} (g (x)) \cdot g^{'} (x)$

When differentiating $y$ , remember to multiply by $\frac{d y}{d x}$

$\begin{array}{rcl} - \sin (x + y) \cdot (\frac{d}{d x} (x + y)) & = & 4 x \\ - \sin (x + y) \cdot (1 + 1 \cdot \frac{d y}{d x}) & = & 4 x \end{array}$

Rearrange to make $\frac{d y}{d x}$ the subject

$1 + \frac{d y}{d x} = \frac{4 x}{- \sin (x + y)} \frac{d y}{d x} = - 1 - \frac{4 x}{\sin (x + y)}$

$\frac{d y}{d x} = - (1 + \frac{4 x}{\sin (x + y)})$

Derivatives of inverse functions using implicit differentiation

How can I differentiate inverse functions using implicit differentiation?

Implicit differentiation provides an alternative method for finding the derivative of inverse functions
Consider differentiating $y = \sin^{- 1} x$
- This can be rewritten as $x = \sin y$
Differentiate using implicit differentiation
- $\frac{d}{d x} (x) = \frac{d}{d x} (\sin y)$
- $1 = \cos y \cdot \frac{d y}{d x}$
Rearrange to make $\frac{d y}{d x}$ the subject
- $\frac{d y}{d x} = \frac{1}{\cos y}$
Use the identity $\sin^{2} y + \cos^{2} y = 1$ , rearranged to make $\cos y$ the subject
- $\cos^{2} y = 1 - \sin^{2} y$
- $\cos y = \pm \sqrt{1 - \sin^{2} y}$
Therefore
- $\frac{d y}{d x} = \pm \frac{1}{\sqrt{1 - \sin^{2} y}}$
But $x = \sin y$ , so
- $\frac{d y}{d x} = \pm \frac{1}{\sqrt{1 - x^{2}}}$

Examiner Tips and Tricks

$\frac{d y}{d x} = \pm \frac{1}{\sqrt{1 - x^{2}}}$ is the general form of $\frac{d y}{d x}$ when $x = \sin y$

When $\sin^{- 1} x$ is defined so as to be the inverse function of $\sin x$ (in which case it is also written as $arc \sin x$ ), its range is restricted to the interval $[- \frac{π}{2}, \frac{π}{2}]$ .

With that restricted range, the derivative of the function is always positive. This is why no ' $\pm$ ' is needed when stating the standard result

$\frac{d}{d x} (\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}}$

A similar method can be used to show why the derivative of $a^{x}$ is $a^{x} \ln a$
- If $y = a^{x}$
  - then $\ln y = x \ln a$
- Differentiate using implicit differentiation
  - Remember that $\frac{d}{d x} (\ln x) = \frac{1}{x}$ and that $\ln a$ is a constant
  - $\frac{1}{y} \cdot \frac{d y}{d x} = \ln a$
- Rearrange
  - $\frac{d y}{d x} = y \ln a$
- Recall that $y = a^{x}$
  - $\frac{d y}{d x} = a^{x} \ln a$

Worked Example

Given that $y = \arctan x$ , use implicit differentiation to show that $\frac{d y}{d x} = \frac{1}{x^{2} + 1}$ .

Answer:

Rewrite the equation in terms of $\tan$ rather than $\arctan$

$\tan y = x$

Differentiate using implicit differentiation

$\tan y$ differentiates to $\sec^{2} y$ , and remember to multiply by $\frac{d y}{d x}$ as we are differentiating with respect to $x$

$\begin{array}{rcl} \frac{d}{d x} (\tan y) & = & \frac{d}{d x} (x) \\ \sec^{2} y \cdot \frac{d y}{d x} & = & 1 \end{array}$

Make $\frac{d y}{d x}$ the subject

$\frac{d y}{d x} = \frac{1}{\sec^{2} y}$

Use the identity $\tan^{2} y + 1 = \sec^{2} y$

$\frac{d y}{d x} = \frac{1}{\tan^{2} y + 1}$

But $\tan y = x$ , so

$\frac{d y}{d x} = \frac{1}{x^{2} + 1}$

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Implicit Differentiation (College Board AP® Calculus BC): Study Guide

Derivatives of implicit functions

What is an implicit function?

What is implicit differentiation?

How do I use implicit differentiation?

How might implicit differentiation questions be made harder?

Derivatives of inverse functions using implicit differentiation

How can I differentiate inverse functions using implicit differentiation?

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Unit 1: Limits & Continuity

Limits

Definition of a Limit

Infinite Limits & Limits at Infinity

Properties of Limits

Evaluating Limits Analytically

Evaluating Limits Numerically & Graphically

Squeeze Theorem & Trigonometric Limits

Summary of Limits

Selecting Procedures for Determining Limits

Continuity

Continuity

Removable Discontinuities

Non-removable Discontinuities

Intermediate Value Theorem

Unit 2: Differentiation Definition & Fundamental Properties

Definition of Differentiation

Average Rate of Change

Instantaneous Rate of Change

Derivatives & Tangents

Estimating the Derivative at a Point

Differentiability & Continuity

Fundamental Properties of Differentiation

Derivative Rules

Derivatives of Exponentials and Logarithms

Derivatives of Sine and Cosine Functions

The Product Rule

The Quotient Rule

Derivatives of Tangent and Reciprocal Trig Functions

Unit 3: Differentiation of Composite, Implicit & Inverse Functions

Differentiation of Composite & Inverse Functions

The Chain Rule

The Inverse Function Theorem

Derivatives of Inverse Trigonometric Functions

Higher-Order Derivatives

Implicit Differentiation

Implicit Differentiation

Summary of Differentiation

Selecting Procedures for Calculating Derivatives

Unit 4: Contextual Applications of Differentiation

Rates of Change & Related Rates

Meaning of a Derivative in Context

Motion in a Straight Line

Related Rates

Linearization

Approximating Values of a Function

L'Hospital's Rule

L'Hospital's Rule

Unit 5: Analytical Applications of Differentiation

Graphs of Functions & Their Derivatives

Mean Value Theorem

Extreme Value Theorem

Critical Points

Increasing & Decreasing Functions

First Derivative Test for Local Extrema

Candidates Test for Global Extrema

Concavity of Functions

Second Derivative Test for Local Extrema

Graphs of f, f' & f''

Optimization Problems

Behaviors of Implicit Relations

Second Derivatives of Implicit Functions

Critical Points of Implicit Relations

Unit 6: Integration & Accumulation of Change

Integration & Antiderivatives

Indefinite Integrals

Derivatives & Antiderivatives

Indefinite Integral Rules

Constant of Integration

Riemann Sums & Definite Integrals