Skills Review for Basics of Differential Equations

Learning Outcomes

  • Apply the chain rule together with the power rule
  • Apply the chain rule and the product/quotient rules correctly in combination when both are necessary
  • Recognize the chain rule for a composition of three or more functions
  • Write function equations using given conditions

In the Basics of Differential Equations section, we will verify the solutions of differential equations by taking derivatives. We will also find exact solutions of differential equations by using given conditions. These skills are reviewed here.

Use the Chain Rule

The Chain Rule

Let [latex]f[/latex] and [latex]g[/latex] be functions. For all [latex]x[/latex] in the domain of [latex]g[/latex] for which [latex]g[/latex] is differentiable at [latex]x[/latex] and [latex]f[/latex] is differentiable at [latex]g(x)[/latex], the derivative of the composite function

[latex]h(x)=(f\circ g)(x)=f(g(x))[/latex]

 

is given by

[latex]h^{\prime}(x)=f^{\prime}(g(x))g^{\prime}(x)[/latex]

 

Alternatively, if [latex]y[/latex] is a function of [latex]u[/latex], and [latex]u[/latex] is a function of [latex]x[/latex], then

[latex]\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}[/latex]

Note that we often need to use the chain rule with other rules. For example, to find derivatives of functions of the form [latex]h(x)=(g(x))^n[/latex], we need to use the chain rule combined with the power rule. To do so, we can think of [latex]h(x)=(g(x))^n[/latex] as [latex]f(g(x))[/latex] where [latex]f(x)=x^n[/latex]. Then [latex]f^{\prime}(x)=nx^{n-1}[/latex]. Thus, [latex]f^{\prime}(g(x))=n(g(x))^{n-1}[/latex]. This leads us to the derivative of a power function using the chain rule,

[latex]h^{\prime}(x)=n(g(x))^{n-1}g^{\prime}(x)[/latex]

Power Rule for Composition of Functions

For all values of [latex]x[/latex] for which the derivative is defined, if

[latex]h(x)=(g(x))^n[/latex]

 

Then

[latex]h^{\prime}(x)=n(g(x))^{n-1}g^{\prime}(x)[/latex]

 

Example: Using the Chain and Power Rules

Find the derivative of [latex]h(x)=\dfrac{1}{(3x^2+1)^2}[/latex]

Try It

Find the derivative of [latex]h(x)=(2x^3+2x-1)^4[/latex]

Example: Using the Chain and Power Rules with a Trigonometric Function

Find the derivative of [latex]h(x)=\sin^3 x[/latex]

Now that we can combine the chain rule and the power rule, we examine how to combine the chain rule with the other rules we have learned. In particular, we can use it with the formulas for the derivatives of trigonometric functions or with the product rule.

Example: Using the Chain Rule on a Cosine Function

Find the derivative of [latex]h(x)= \cos (5x^2)[/latex].

Example: Using the Chain Rule on Another Trigonometric Function

Find the derivative of [latex]h(x)= \sec (4x^5+2x)[/latex].

Try It

Find the derivative of [latex]h(x)= \sin (7x+2)[/latex].

We now provide a list of derivative formulas that may be obtained by applying the chain rule in conjunction with the formulas for derivatives of trigonometric functions. Their derivations are similar to those used in the last three examples. For convenience, formulas are also given in Leibniz’s notation, which some students find easier to remember. (We discuss the chain rule using Leibniz’s notation at the end of this section.) It is not absolutely necessary to memorize these as separate formulas as they are all applications of the chain rule to previously learned formulas.

Using the Chain Rule with Trigonometric Functions

For all values of [latex]x[/latex] for which the derivative is defined,

[latex]\begin{array}{llll}\frac{d}{dx}(\sin (g(x)))= \cos (g(x))g^{\prime}(x) & & & \frac{d}{dx} \sin u= \cos u\frac{du}{dx} \\ \frac{d}{dx}(\cos (g(x)))=−\sin (g(x))g^{\prime}(x) & & & \frac{d}{dx} \cos u=−\sin u\frac{du}{dx} \\ \frac{d}{dx}(\tan (g(x)))= \sec^2 (g(x))g^{\prime}(x) & & & \frac{d}{dx} \tan u=\sec^2 u\frac{du}{dx} \\ \frac{d}{dx}(\cot (g(x)))=−\csc^2 (g(x))g^{\prime}(x) & & & \frac{d}{dx} \cot u=−\csc^2 u\frac{du}{dx} \\ \frac{d}{dx}(\sec (g(x)))= \sec (g(x)) \tan (g(x))g^{\prime}(x) & & & \frac{d}{dx} \sec u= \sec u \tan u\frac{du}{dx} \\ \frac{d}{dx}(\csc (g(x)))=−\csc (g(x)) \cot (g(x))g^{\prime}(x) & & & \frac{d}{dx} \csc u=−\csc u \cot u\frac{du}{dx} \end{array}[/latex]

Write Function Equations Using Given Conditions

Sometimes, to find a missing value in a function equation, you will be given an input of the function and the corresponding output. You will then plug this input and output into the function equation and find the missing value.

Example: Writing a Function Equation from given conditions

Given [latex]f(2)=-1[/latex], find the unknown value c in the function equation [latex]f(x)=3x^3-4x^2-x+c[/latex].

 

Try It

Given [latex]f(1)=5[/latex], find the unknown value c in the function equation [latex]f(x)=-2x^2+3x+c[/latex].