Second Derivatives of Implicit Functions (College Board AP® Calculus AB)

Study Guide

Written by: Jamie Wood

Reviewed by: Dan Finlay

Second derivatives of implicit functions

How do I find the second derivative of an implicit function?

First make sure you are comfortable with the content covered in the 'Implicit Differentiation' study guide
- This involves being able to find the derivative of a function defined implicitly, using the chain rule
We are able to find the second derivative, $\frac{d^{2} y}{d x^{2}}$ , by differentiating the expression for the first derivative, $\frac{d y}{d x}$
Consider finding the second derivative of $x^{2} + y^{2} = 4 x$
Find the first derivative, this is shown in the Implicit Differentiation study guide to be
- $\frac{d y}{d x} = \frac{2 - x}{y}$
To find the second derivative, differentiate both sides with respect to $x$
- $\frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\frac{2 - x}{y})$
The derivative of $\frac{d y}{d x}$ with respect to $x$ is $\frac{d^{2} y}{d x^{2}}$
- $\frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{2 - x}{y})$
The right hand side can then be differentiated with respect to $x$ , in this case using the quotient rule, ${(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$
- $u = 2 - x$ and $v = y$
Differentiate $u$ and $v$ with respect to $x$ , remember to apply the chain rule when differentiating $y$
- $u^{'} = - 1$ and $v^{'} = \frac{d y}{d x}$
Applying the quotient rule to the right hand side
- $\frac{d^{2} y}{d x^{2}} = \frac{- 1 \cdot y - (2 - x) \cdot \frac{d y}{d x}}{y^{2}}$
If your answer references the first derivative, substitute it in
- We know from before that $\frac{d y}{d x} = \frac{2 - x}{y}$
- $\frac{d^{2} y}{d x^{2}} = \frac{- 1 \cdot y - (2 - x) \cdot (\frac{2 - x}{y})}{y^{2}}$
This is the correct second derivative, but it can be simplified
- $\frac{d^{2} y}{d x^{2}} = \frac{- y - \frac{{(2 - x)}^{2}}{y}}{y^{2}}$
Writing with a common denominator can help simplify
- $\frac{d^{2} y}{d x^{2}} = \frac{- y}{y^{2}} - \frac{{(2 - x)}^{2}}{y^{3}} = \frac{- y^{2}}{y^{3}} - \frac{{(2 - x)}^{2}}{y^{3}}$
$\frac{d^{2} y}{d x^{2}} = - \frac{y^{2} + {(2 - x)}^{2}}{y^{3}}$
This is the final answer, but expressions like this can be written in several different ways,
- E.g. by factoring out negative signs in a different place

Examiner Tips and Tricks

Remember that:

The derivative of $y$ with respect to $x$ is $\frac{d y}{d x}$
The derivative of $\frac{d y}{d x}$ with respect to $x$ is $\frac{d^{2} y}{d x^{2}}$

Worked Example

Show that the second derivative of $\sin x + \cos y = 0.8$ can be written as:

$\frac{d^{2} y}{d x^{2}} = - \frac{\sin x \sin^{2} y + \cos^{2} x \cos y}{\sin^{3} y}$

Answer:

Find the first derivative by differentiating both sides with respect to $x$

$\begin{array}{rcl} \frac{d}{d x} (\sin x + \cos y) & = & \frac{d}{d x} (0.8) \\ \frac{d}{d x} \sin x + \frac{d}{d x} \cos y & = & \frac{d}{d x} (0.8) \end{array}$

Remember to use the chain rule when differentiating $\cos y$ with respect to $x$

$\begin{array}{rcl} \cos x + - \sin y \cdot \frac{d y}{d x} & = & 0 \\ \frac{d y}{d x} & = & \frac{\cos x}{\sin y} \end{array}$

This could instead be simplified to $\cos x \csc y$ , depending on whether you would prefer to apply the quotient rule to $\frac{\cos x}{\sin y}$ or the product rule to $\cos x \csc y$ for the next part

To find the second derivative, differentiate both sides with respect to $x$ again

$\begin{array}{rcl} \frac{d}{d x} (\frac{d y}{d x}) & = & \frac{d}{d x} (\frac{\cos x}{\sin y}) \\ \frac{d^{2} y}{d x^{2}} & = & \frac{d}{d x} (\frac{\cos x}{\sin y}) \end{array}$

Apply the quotient rule to the right hand side, remember to apply the chain rule when differentiating $\sin y$ with respect to $x$

${(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$

$u = \cos x$ $v = \sin y$

$u^{'} = - \sin x$ $v^{'} = \cos y \cdot \frac{d y}{d x}$

$\frac{d^{2} y}{d x^{2}} = \frac{- \sin x \cdot \sin y - \cos x \cdot \cos y \cdot \frac{d y}{d x}}{\sin^{2} y}$

Substitute in the expression for $\frac{d y}{d x}$ found previously, $\begin{array}{rcl} \frac{d y}{d x} & = & \frac{\cos x}{\sin y} \end{array}$

$\frac{d^{2} y}{d x^{2}} = \frac{- \sin x \cdot \sin y - \cos x \cdot \cos y \cdot (\frac{\cos x}{\sin y})}{\sin^{2} y}$

Now we need to rearrange into the form given in the question

Write as two fractions to see if they can be simplified

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

They can be written with a common denominator

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

Put the negative at the front of the entire fraction, to match the form given in the question

$\frac{d^{2} y}{d x^{2}} = - \frac{\sin x \sin^{2} y + \cos^{2} x \cos y}{\sin^{3} y}$

Last updated: 21 August 2024

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Previous:Optimization ProblemsNext:Critical Points of Implicit Relations

Second Derivatives of Implicit Functions (College Board AP® Calculus AB)

Study Guide

Second derivatives of implicit functions

How do I find the second derivative of an implicit function?

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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

Continuous Functions

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

Accumulation of Change

Riemann Sums