Implicit Differentiation (College Board AP® Calculus AB)

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Written by: Jamie Wood

Reviewed by: Dan Finlay

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
- $\frac{d y}{d x} = \frac{1}{\cos y}$
Recall that $y = \sin^{- 1} x$
- $\frac{d y}{d x} = \frac{1}{\cos (\sin^{- 1} x)}$
Use the identity $\sin^{2} x + \cos^{2} x = 1$ rearranged to $\cos x = \sqrt{1 - \sin^{2} x}$
- $\frac{d y}{d x} = \frac{1}{\sqrt{1 - \sin^{2} (\sin^{- 1} x)}}$
- $\frac{d y}{d x} = \frac{1}{\sqrt{1 - x^{2}}}$
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

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

Recall that $y = \arctan x$

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

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

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

Simplify

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

Last updated: 18 September 2024

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Implicit Differentiation (College Board AP® Calculus AB)

Study Guide

Derivatives of implicit functions

What is an implicit function?

What is implicit differentiation?

How do I use implicit differentiation?

Worked Example

How might implicit differentiation questions be made harder?

Worked Example

Worked Example

Derivatives of inverse functions using implicit differentiation

How can I differentiate inverse functions using implicit differentiation?

Worked Example

You've read 0 of your 5 free study guides this week

Sign up now. It’s free!

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

Author: Jamie Wood

Author: Dan Finlay