Skills Review for the Chain Rule

Learning Outcomes

Find the derivative of an implicit function by using implicit differentiation
Write the equation of a line using slope and a point on the line

In the Chain Rule section, we will learn how to apply the chain rule to functions of several variables. Here we will review implicit differentiation and how to write the equation of a line.

What is Implicit Differentiation?

Implicit differentiation allows us to find slopes of tangents to curves that are clearly not functions (they fail the vertical line test). We are using the idea that portions of $y$ are functions that satisfy the given equation, but that $y$ is not actually a function of $x$ .

To perform implicit differentiation on an equation that defines a function $y$ implicitly in terms of a variable $x$ , use the following steps:

Take the derivative of both sides of the equation. Keep in mind that $y$ is a function of $x$ . Consequently, whereas $\frac{d}{dx}(\sin x)= \cos x, \, \frac{d}{dx}(\sin y)= \cos y\frac{dy}{dx}$ because we must use the Chain Rule to differentiate $\sin y$ with respect to $x$ .
Rewrite the equation so that all terms containing $\frac{dy}{dx}$ are on the left and all terms that do not contain $\frac{dy}{dx}$ are on the right.
Factor out $\frac{dy}{dx}$ on the left.
Solve for $\frac{dy}{dx}$ by dividing both sides of the equation by an appropriate algebraic expression.

Example: Using Implicit Differentiation

Assuming that $y$ is defined implicitly by the equation $x^2+y^2=25$ , find $\frac{dy}{dx}$ .

Show Solution

Follow the steps in the problem-solving strategy.

\begin{array}{llll} \frac{d}{d x} (x^{2} + y^{2}) = \frac{d}{d x} (25) & Step 1. Differentiate both sides of the equation. \\ \frac{d}{d x} (x^{2}) + \frac{d}{d x} (y^{2}) = 0 & \begin{array}{l} Step 1.1. Use the sum rule on the left. \\ On the right, \frac{d}{d x} (25) = 0. \end{array} \\ 2 x + 2 y \frac{d y}{d x} = 0 & \begin{array}{l} Step 1.2. Take the derivatives, so \frac{d}{d x} (x^{2}) = 2 x \\ and \frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x} . \end{array} \\ 2 y \frac{d y}{d x} = - 2 x & \begin{array}{l} Step 2. Keep the terms with \frac{d y}{d x} on the left. \\ Move the remaining terms to the right. \end{array} \\ \frac{d y}{d x} = - \frac{x}{y} & \begin{array}{l} Step 4. Divide both sides of the equation by \\ 2 y . (Step 3 does not apply in this case.) \end{array} \end{array}

$\begin{array}{llll} \frac{d}{dx}(x^2+y^2) = \frac{d}{dx}(25) & & & \text{Step 1. Differentiate both sides of the equation.} \\ \frac{d}{dx}(x^2)+\frac{d}{dx}(y^2) = 0 & & & \begin{array}{l}\text{Step 1.1. Use the sum rule on the left.} \\ \text{On the right,} \, \frac{d}{dx}(25)=0. \end{array} \\ 2x+2y\frac{dy}{dx} = 0 & & & \begin{array}{l}\text{Step 1.2. Take the derivatives, so} \, \frac{d}{dx}(x^2)=2x \\ \text{and} \, \frac{d}{dx}(y^2)=2y\frac{dy}{dx}. \end{array} \\ 2y\frac{dy}{dx} = -2x & & & \begin{array}{l}\text{Step 2. Keep the terms with} \, \frac{dy}{dx} \, \text{on the left.} \\ \text{Move the remaining terms to the right.} \end{array} \\ \frac{dy}{dx} = -\frac{x}{y} & & & \begin{array}{l}\text{Step 4. Divide both sides of the equation by} \\ 2y. \, \text{(Step 3 does not apply in this case.)} \end{array} \end{array}$

Analysis

Note that the resulting expression for $\frac{dy}{dx}$ is in terms of both the independent variable $x$ and the dependent variable $y$ . Although in some cases it may be possible to express $\frac{dy}{dx}$ in terms of $x$ only, it is generally not possible to do so.

Example: Using Implicit Differentiation and the Product Rule

Assuming that $y$ is defined implicitly by the equation $x^3 \sin y+y=4x+3$ , find $\frac{dy}{dx}$ .

Show Solution

\begin{array}{llll} \frac{d}{d x} (x^{3} \sin y + y) = \frac{d}{d x} (4 x + 3) & Step 1: Differentiate both sides of the equation. \\ \frac{d}{d x} (x^{3} \sin y) + \frac{d}{d x} (y) = 4 & \begin{array}{l} Step 1.1: Apply the sum rule on the left. \\ On the right, \frac{d}{d x} (4 x + 3) = 4. \end{array} \\ (\frac{d}{d x} (x^{3}) \cdot \sin y + \frac{d}{d x} (\sin y) \cdot x^{3}) + \frac{d y}{d x} = 4 & \begin{array}{l} Step 1.2: Use the product rule to find \\ \frac{d}{d x} (x^{3} \sin y) . Observe that \frac{d}{d x} (y) = \frac{d y}{d x} . \end{array} \\ 3 x^{2} \sin y + (\cos y \frac{d y}{d x}) \cdot x^{3} + \frac{d y}{d x} = 4 & \begin{array}{l} Step 1.3: We know \frac{d}{d x} (x^{3}) = 3 x^{2} . Use the \\ chain rule to obtain \frac{d}{d x} (\sin y) = \cos y \frac{d y}{d x} . \end{array} \\ x^{3} \cos y \frac{d y}{d x} + \frac{d y}{d x} = 4 - 3 x^{2} \sin y & \begin{array}{l} Step 2: Keep all terms containing \frac{d y}{d x} on the \\ left. Move all other terms to the right. \end{array} \\ \frac{d y}{d x} (x^{3} \cos y + 1) = 4 - 3 x^{2} \sin y & Step 3: Factor out \frac{d y}{d x} on the left. \\ \frac{d y}{d x} = \frac{4 - 3 x^{2} \sin y}{x^{3} \cos y + 1} & \begin{array}{l} Step 4: Solve for \frac{d y}{d x} by dividing both sides of \\ the equation by x^{3} \cos y + 1. \end{array} \end{array}

$\begin{array}{llll}\frac{d}{dx}(x^3 \sin y+y) = \frac{d}{dx}(4x+3) & & & \text{Step 1: Differentiate both sides of the equation.} \\ \frac{d}{dx}(x^3 \sin y)+\frac{d}{dx}(y) = 4 & & & \begin{array}{l}\text{Step 1.1: Apply the sum rule on the left.} \\ \text{On the right,} \, \frac{d}{dx}(4x+3)=4. \end{array} \\ (\frac{d}{dx}(x^3) \cdot \sin y+\frac{d}{dx}(\sin y) \cdot x^3) + \frac{dy}{dx} = 4 & & & \begin{array}{l}\text{Step 1.2: Use the product rule to find} \\ \frac{d}{dx}(x^3 \sin y). \, \text{Observe that} \, \frac{d}{dx}(y)=\frac{dy}{dx}. \end{array} \\ 3x^2 \sin y+(\cos y\frac{dy}{dx}) \cdot x^3 + \frac{dy}{dx} = 4 & & & \begin{array}{l}\text{Step 1.3: We know} \, \frac{d}{dx}(x^3)=3x^2. \, \text{Use the} \\ \text{chain rule to obtain} \, \frac{d}{dx}(\sin y)= \cos y\frac{dy}{dx}. \end{array} \\ x^3 \cos y\frac{dy}{dx}+\frac{dy}{dx} = 4-3x^2 \sin y & & & \begin{array}{l}\text{Step 2: Keep all terms containing} \, \frac{dy}{dx} \, \text{on the} \\ \text{left. Move all other terms to the right.} \end{array} \\ \frac{dy}{dx}(x^3 \cos y+1) = 4-3x^2 \sin y & & & \text{Step 3: Factor out} \, \frac{dy}{dx} \, \text{on the left.} \\ \frac{dy}{dx} = \large \frac{4-3x^2 \sin y}{x^3 \cos y+1} & & & \begin{array}{l}\text{Step 4: Solve for} \, \frac{dy}{dx} \, \text{by dividing both sides of} \\ \text{the equation by} \, x^3 \cos y+1. \end{array} \end{array}$

Try It

Find $\frac{dy}{dx}$ for $y$ defined implicitly by the equation $4x^5+ \tan y=y^2+5x$ .

Hint

Follow the problem solving strategy, remembering to apply the chain rule to differentiate $\tan y$ and $y^2$ .

Show Solution

$\frac{dy}{dx}=\large \frac{5-20x^4}{\sec^2 y-2y}$

Try It

Write the Equation of a Line

(also in Module 1, Skills Review for Parametric Equations and Calculus of Parametric Equations)
The slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between any two points on a line.

A General Note: The Slope of a Line

The slope of a line, m, represents the change in y over the change in x. Given two points, $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$ , the following formula determines the slope of a line containing these points:

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

$m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$

Example: Finding the Slope of a Line Given Two Points

Find the slope of a line that passes through the points $\left(2,-1\right)$ and $\left(-5,3\right)$ .

Show Solution

We substitute the y-values and the x-values into the formula.

\begin{array}{l} m & = \frac{3 - (- 1)}{- 5 - 2} \\ = \frac{4}{- 7} \\ = - \frac{4}{7} \end{array}

$\begin{array}{l}m\hfill&=\frac{3-\left(-1\right)}{-5 - 2}\hfill \\ \hfill&=\frac{4}{-7}\hfill \\ \hfill&=-\frac{4}{7}\hfill \end{array}$

The slope is $-\frac{4}{7}$ .

Analysis of the Solution

It does not matter which point is called $\left({x}_{1},{y}_{1}\right)$ or $\left({x}_{2},{y}_{2}\right)$ . As long as we are consistent with the order of the y terms and the order of the x terms in the numerator and denominator, the calculation will yield the same result.

Try It

Find the slope of the line that passes through the points $\left(-2,6\right)$ and $\left(1,4\right)$ .

Show Solution

slope $=m=\dfrac{-2}{3}=-\dfrac{2}{3}$

To write the equation of a line, the line’s slope and a point the line goes through must be known. Perhaps the most familiar form of a linear equation is slope-intercept form written as $y=mx+b$ , where $m=\text{slope}$ and $b=y\text{-intercept}$ . Let us begin with the slope.

Often, the starting point to writing the equation of a line is to use point-slope formula. Given the slope and one point on a line, we can find the equation of the line using point-slope form shown below.

y - y_{1} = m (x - x_{1})

$y-{y}_{1}=m\left(x-{x}_{1}\right)$

We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.

A General Note: The Point-Slope Formula

Given one point and the slope, using point-slope form will lead to the equation of a line:

y - y_{1} = m (x - x_{1})

$y-{y}_{1}=m\left(x-{x}_{1}\right)$

Example: Finding the Equation of a Line Given the Slope and One Point

Write the equation of the line with slope $m=-3$ and passing through the point $\left(4,8\right)$ . Write the final equation in slope-intercept form.

Show Solution

Using point-slope form, substitute $-3$ for m and the point $\left(4,8\right)$ for $\left({x}_{1},{y}_{1}\right)$ .

\begin{array}{l} y - y_{1} = m (x - x_{1}) \\ y - 8 = - 3 (x - 4) \\ y - 8 = - 3 x + 12 \\ y = - 3 x + 20 \end{array}

$\begin{array}{l}y-{y}_{1}=m\left(x-{x}_{1}\right)\hfill \\ y - 8=-3\left(x - 4\right)\hfill \\ y - 8=-3x+12\hfill \\ y=-3x+20\hfill \end{array}$

Analysis of the Solution

Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.

Try It

Given $m=4$ , find the equation of the line in slope-intercept form passing through the point $\left(2,5\right)$ .

Show Solution

$y=4x - 3$

Differentiation of Functions of Several Variables: Prerequisite Material