Skills Review for Taylor and Maclaurin Series and Working with Taylor Series

Learning Outcomes

Apply the power rule
Find the derivatives of the sine and cosine function.
Find the derivatives of the standard trigonometric functions.
Find the derivative of exponential functions
Find the derivative of logarithmic functions
Apply factorial notation

In the Taylor and Maclaurin Series and Working with Taylor Series sections, we will look at how to construct Taylor and Maclaurin polynomials. Here we will review how to take derivatives of various types of functions and how to use factorial notation.

Apply the Power Rule

(also in Module 5, Skills Review for Integration by Parts)

We know that

\frac{d}{d x} (x^{2}) = 2 x

$\dfrac{d}{dx}\left(x^2\right)=2x$ and

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

$\dfrac{d}{dx}\left(x^{\frac{1}{2}}\right)=\dfrac{1}{2}x^{−\frac{1}{2}}$

As we shall see, there is a procedure for finding the derivative of the general form $f(x)=x^n$ . The following theorem states that this power rule holds for all non-variable powers of $x$ .

The Power Rule

Let $n$ be a number. If $f(x)=x^n$ , then

f^{'} (x) = n x^{n - 1}

$f^{\prime}(x)=nx^{n-1}$

Alternatively, we may express this rule as

\frac{d}{d x} (x^{n}) = n x^{n - 1}

$\dfrac{d}{dx}(x^n)=nx^{n-1}$

Example: Applying Basic Derivative Rules

Find the derivative of the function $f(x)=x^{10}$ by applying the power rule.

Show Solution

Using the power rule with $n=10$ , we obtain

f^{'} (x) = 10 x^{10 - 1} = 10 x^{9}

$f^{\prime}(x)=10x^{10-1}=10x^9$ .

Example: Applying Basic Derivative Rules

Find the derivative of $f(x)=2x^5+7$ .

Show Solution

We begin by applying the rule for differentiating the sum of two functions, followed by the rules for differentiating constant multiples of functions and the rule for differentiating powers. To better understand the sequence in which the differentiation rules are applied, we use Leibniz notation throughout the solution:

\begin{array}{lllll} f^{'} (x) & = \frac{d}{d x} (2 x^{5} + 7) \\ = \frac{d}{d x} (2 x^{5}) + \frac{d}{d x} (7) & Take the derivative term by term. \\ = 2 \frac{d}{d x} (x^{5}) + \frac{d}{d x} (7) & Apply the constant multiple rule. \\ = 2 (5 x^{4}) + 0 & Apply the power rule and the constant rule. \\ = 10 x^{4} . & Simplify. \end{array}

$\begin{array}{lllll}f^{\prime}(x) & =\frac{d}{dx}(2x^5+7) & & & \\ & =\frac{d}{dx}(2x^5)+\frac{d}{dx}(7) & & & \text{Take the derivative term by term.} \\ & =2\frac{d}{dx}(x^5)+\frac{d}{dx}(7) & & & \text{Apply the constant multiple rule.} \\ & =2(5x^4)+0 & & & \text{Apply the power rule and the constant rule.} \\ & =10x^4. & & & \text{Simplify.} \end{array}$

Example: Applying Basic Derivative Rules

Find the derivative of $f(x)=\sqrt{x}$ .

Show Solution

For any function in radical form, rewrite it in exponential form.

$f(x)=\sqrt{x}$ can be rewritten as $f(x)=x^\frac{1}{2}$ .

Now, take the derivative using the power rule: $f^{\prime}(x)=\dfrac{1}{2}x^\frac{-1}{2}$ .

Try It

Find the derivative of $f(x)=2x^{-3}-6x^2+3$ .

Show Solution

$f^{\prime}(x)=-6x^{-4}-12x$ .

Try It

Find the derivative of $f(x)=\sqrt{x^7}$ .

Show Solution

Rewrite: $f(x)=x^\frac{7}{2}$

Differentiate: $f^{\prime}(x)=\dfrac{7}{2}x^\frac{5}{2}$

Find the Derivatives of the Standard Trigonometric Functions.

(also in Module 5, Skills Review for Integration by Parts)

The Derivatives of $\sin x$ and $\cos x$

The derivative of the sine function is the cosine and the derivative of the cosine function is the negative sine.

\frac{d}{d x} (\sin x) = \cos x

$\frac{d}{dx}(\sin x)= \cos x$

\frac{d}{d x} (\cos x) = - \sin x

$\frac{d}{dx}(\cos x)=−\sin x$

Example: Differentiating a Function Containing $sinx$

Find the derivative of $f(x)=5x^3 \sin x$ .

Show Solution

Using the product rule, we have

\begin{array}{ll} f^{'} (x) & = \frac{d}{d x} (5 x^{3}) \cdot \sin x + \frac{d}{d x} (\sin x) \cdot 5 x^{3} \\ = 15 x^{2} \cdot \sin x + \cos x \cdot 5 x^{3} \end{array}

$\begin{array}{ll}f^{\prime}(x) & =\frac{d}{dx}(5x^3)\cdot \sin x+\frac{d}{dx}(\sin x)\cdot 5x^3 \\ & =15x^2\cdot \sin x+ \cos x\cdot 5x^3\end{array}$

After simplifying, we obtain

f^{'} (x) = 15 x^{2} \sin x + 5 x^{3} \cos x

$f^{\prime}(x)=15x^2 \sin x+5x^3 \cos x$ .

The Derivatives of $\tan x, \, \cot x, \, \sec x$ , and $\csc x$

The derivatives of the remaining trigonometric functions are as follows:

\frac{d}{d x} (\tan x) = \sec^{2} x

$\frac{d}{dx}(\tan x)=\sec^2 x$

\frac{d}{d x} (\cot x) = - \csc^{2} x

$\frac{d}{dx}(\cot x)=−\csc^2 x$

\frac{d}{d x} (\sec x) = \sec x \tan x

$\frac{d}{dx}(\sec x)= \sec x \tan x$

\frac{d}{d x} (\csc x) = - \csc x \cot x

$\frac{d}{dx}(\csc x)=−\csc x \cot x$

Try It

Find the derivative of $f(x)= \cot x$ .

Show Solution

$f^{\prime}(x)=−\csc^2 x$

Find the Derivatives of Exponential and Logarithmic Functions with Base e

(also in Module 5, Skills Review for Integration by Parts)

Derivative of the Natural Exponential Function

Let $E(x)=e^x$ be the natural exponential function. Then

E^{'} (x) = e^{x}

$E^{\prime}(x)=e^x$

In general,

\frac{d}{d x} (e^{g (x)}) = e^{g (x)} g^{'} (x)

$\frac{d}{dx}(e^{g(x)})=e^{g(x)} g^{\prime}(x)$

If it helps, think of the formula as the chain rule being applied to natural exponential functions. The derivative of $e$ raised to the power of a function will simply be $e$ raised to the power of the function multiplied by the derivative of that function.

Example: Differentiating An Exponential Function

Find the derivative of $f(x)=e^{\tan (2x)}$ .

Show Solution

Using the derivative formula and the chain rule,

$\begin{array}{ll} f^{\prime}(x) & =e^{\tan (2x)}\frac{d}{dx}(\tan (2x)) \\ & = e^{\tan (2x)} \sec^2 (2x) \cdot 2. \end{array}$

Recall that for exponential functions, we apply the chain rule to the exponent.

Try It

Find the derivative of $f(x)=e^{5x^2}$ .

Show Solution

$f^{\prime}(x)=10xe^{5x^2}$

Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic function.

The Derivative of the Natural Logarithmic Function

If $x>0$ and $y=\ln x$ , then

\frac{d y}{d x} = \frac{1}{x}

$\frac{dy}{dx}=\dfrac{1}{x}$

More generally, let $g(x)$ be a differentiable function. For all values of $x$ for which $g^{\prime}(x)>0$ , the derivative of $h(x)=\ln(g(x))$ is given by

h^{'} (x) = \frac{1}{g (x)} g^{'} (x)

$h^{\prime}(x)=\dfrac{1}{g(x)} g^{\prime}(x)$

Example: Differentiating A Natural Logarithmic Function

Find the derivative of $f(x)=\ln(x^3+3x-4)$

Show Solution

Use the derivative of a natural logarithm directly.

\begin{array}{lllll} f^{'} (x) & = \frac{1}{x^{3} + 3 x - 4} \cdot (3 x^{2} + 3) & Use g (x) = x^{3} + 3 x - 4 in h^{'} (x) = \frac{1}{g (x)} g^{'} (x) . \\ = \frac{3 x^{2} + 3}{x^{3} + 3 x - 4} & Rewrite. \end{array}

$\begin{array}{lllll} f^{\prime}(x) & =\frac{1}{x^3+3x-4} \cdot (3x^2+3) & & & \text{Use} \, g(x)=x^3+3x-4 \, \text{in} \, h^{\prime}(x)=\frac{1}{g(x)} g^{\prime}(x). \\ & =\frac{3x^2+3}{x^3+3x-4} & & & \text{Rewrite.} \end{array}$

Try It

Find the derivative of $g(x)=\ln(3x+7)$ .

Show Solution

$g^{\prime}(x)=\dfrac{3}{3x+7}$

Apply Factorial Notation

(See Module 6, Skills Review for Power Series and Functions)

Power Series: Prerequisite Material