Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 4

Differentiation

Derivatives of the Trigonometric Functions

Before we find the derivative functions of the trigonometric functions, we should try to find their graphs. See how this is done for the sine function on page 145 of the textbook. We will do the same for the cosine function.

The graphs of cosine and its derivative function are shown in Figure 4.17, below.

Figure 4.17. Cosine and its derivative function

We suspect from these graphs that the derivative functions of the sine and cosine are “wave” functions.

On page 145 of the textbook, you can see how Definition 4.5 is applied to find the derivative of the sine function. The derivative of the cosine function is obtained in the same manner.

${\displaystyle \begin{array}{lll}\hfill {\displaystyle \frac{d}{dx}}\cos x& =\underset{h\to 0}{\lim }{\displaystyle \frac{\cos \left(x+h\right)-\cos \left(x\right)}{h}}\hfill & \text{Definition 4.5}\hfill \\ \hfill & =\underset{h\to 0}{\lim }{\displaystyle \frac{\cos x\mathrm{ cos}h-\mathrm{sin }x\mathrm{ sin}h-\cos x}{h}}\hfill & \text{addition formula}\hfill \\ \hfill & =\underset{h\to 0}{\lim }\mathrm{ cos}x\left({\displaystyle \frac{\cos h-1}{h}}\right)-\sin x\left({\displaystyle \frac{\sin h}{h}}\right)\hfill & \text{factoring}\hfill \\ \hfill & =\cos x\underset{h\to 0}{\lim }{\displaystyle \frac{\cos h-1}{h}}-\sin x\underset{h\to 0}{\lim }{\displaystyle \frac{\sin h}{h}}\hfill & \text{laws of limits}\hfill \\ \hfill & =\cos x\left(0\right)-\sin x\left(1\right)\hfill & \hfill \\ \hfill & =-\sin x\text{.}\hfill & \hfill \end{array}}$

You can see on page 148 of the textbook how the quotient rule is applied to obtain the derivative of

${\displaystyle \mathrm{tan }x={\displaystyle \frac{\mathrm{sin }x}{\mathrm{cos }x}}\text{.}}$

We can use the reciprocal rule to obtain the derivative of the cotangent function. If

${\displaystyle {\displaystyle \frac{d}{dx}}\mathrm{ tan }x={\sec }^{2}x\text{,}}$

then

Similarly, the reciprocal rule is used to obtain the derivatives of the secant and cosecant functions. All the derivative functions of the trigonometric functions are listed on page 148 of the textbook. You must know all of them. The effective way to memorize them is by doing many exercises.

It should be clear to you that the rules of differentiation can be applied to differentiate the sum, product and quotient of more than two functions.

Example 4.35. The function $f\left(x\right)=x^2 sin x\left(\sqrt{x}+5\right)$ is the product of three functions. [Which ones?] To apply the product rule, we consider it as a product of two functions, $f\left(x\right)=\left(x^2 sin x\right)\left(\sqrt{x}+5\right)$, and apply the product rule to these two functions. We first find the derivative of $g\left(x\right)=x^2 sin x$, again using the product rule.

$g^{\prime }\left(x\right)=2x\mathrm{ sin }x+x^2\mathrm{cos }x\text{,}$

and

${\displaystyle {\displaystyle \frac{d}{dx}}\sqrt{x}+5={\displaystyle \frac{1}{2\sqrt{x}}}\text{.}}$

So,

${\displaystyle \begin{array}{ll}\hfill f^{\prime }\left(x\right)& =\left[{\displaystyle \frac{d}{dx}}\left(x^2\sin x\right)\right]\left(\sqrt{x}+5\right)+\left(x^2\sin x\right){\displaystyle \frac{d}{dx}}\left(\sqrt{x}+5\right)\hfill \\ \hfill & =\left(2x\sin x+x^2\cos x\right)\left(\sqrt{x}+5\right)+\left(x^2\sin x\right)\left({\displaystyle \frac{1}{2\sqrt{x}}}\right)\hfill \\ \hfill & =\left({\displaystyle \frac{5}{2}}\sqrt{x}+10\right)x\sin x+\left(\sqrt{x}+5\right)x^2\cos x\text{.}\hfill \end{array}}$

Example 4.36. When we differentiate, we should simplify the function (if possible) before applying the rules of differentiation. To differentiate the function $f\left(x\right)=\mathrm{cot }x{ sec}^{\mathrm{2 }}x$ we simplify, obtaining

${\displaystyle f\left(x\right)={\displaystyle \frac{\mathrm{cos }x}{\mathrm{sin }x}}{\displaystyle \frac{1}{{cos}^{2}x}}={\displaystyle \frac{1}{\mathrm{sin }x\mathrm{ cos }x}}\text{.}}$

We find the derivative of $g\left(x\right)=\mathrm{sin }x cos x$ using the product rule.

${\displaystyle \begin{array}{lll}{\displaystyle \frac{d}{dx}}\sin x\cos x\hfill & =\left({\displaystyle \frac{d}{dx}}\sin x\right)\cos x+\sin x\left({\displaystyle \frac{d}{dx}}\cos x\right)\hfill & \text{product rule}\hfill \\ \hfill & =\cos x\left(\cos x\right)+\left(\sin x\right)\left(-\sin x\right)\hfill & \hfill \\ \hfill & ={\cos }^{2}x-{\sin }^{2}x\text{.}\hfill & \hfill \end{array}}$

Applying the reciprocal rule,

${\displaystyle \begin{array}{lll}\hfill {\displaystyle \frac{d}{dx}}{\displaystyle \frac{1}{\sin x\cos x}}& =-{\displaystyle \frac{{\cos }^{2}x-{\sin }^{2}x}{{\left(\sin x\cos x\right)}^2}}\hfill & \text{reciprocal rule}\hfill \\ \hfill & =-{\displaystyle \frac{{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}}+{\displaystyle \frac{{\sin }^{2}x}{{\sin }^{2}x{\cos }^{2}x}}\hfill & \text{simplifying}\hfill \\ \hfill & =-{\displaystyle \frac{1}{{\sin }^{2}x}}+{\displaystyle \frac{1}{{\cos }^{2}x}}\hfill & \hfill \\ \hfill & =-{\csc }^{2}x+{\sec }^{2}x\text{.}\hfill & \hfill \end{array}}$

You may want to apply the product rule to the function $f\left(x\right)=\mathrm{cot }x{\mathrm{ sec}}^{\mathrm{2 }}x$ and compare the methods to see which procedure is easier.

Example 4.37. To differentiate the function

${\displaystyle f\left(x\right)={\displaystyle \frac{x^4\mathrm{tan }x+\mathrm{cos }x}{x sin x+3}}\text{.}}$

we start by deciding which differentiation rule we need to apply first. In this case, we will need the quotient rule, but to apply it, we must first find the derivatives of the numerator and denominator functions. For the derivative of the functions $g\left(x\right)=x^4\mathrm{ tan }x+\mathrm{cos }x$ and $h\left(x\right)=x\mathrm{ sin }x+3$, we apply the product and sum rules

$\begin{array}{rll}{g}^{\prime }(x)& =\left({\displaystyle \frac{d}{dx}}{x}^4\right)\tan x+x^4\left({\displaystyle \frac{d}{dx}}\tan x\right)+{\displaystyle \frac{d}{dx}}\cos x& \text{sum}\text{and}\text{product}\text{rules}\\ & =4x^3\tan x+x^4{\sec }^{2}x-\sin x& \text{power}\text{rule}\\ & =x^3(4\tan x+x{\sec }^{2}x)-\sin x& \end{array}$

and

${\displaystyle h^{\prime }\left(x\right)=\left({\displaystyle \frac{d}{dx}}x\right)\left(\mathrm{sin }x\right)+x\left({\displaystyle \frac{d}{dx}}\mathrm{sin }x\right)= sin x+x cos x\text{.}}$

Now, we can apply the quotient rule:

Observe that by applying the quotient rule step by step, we reduce the possibility of mistakes.

Example 4.38. For the function $f\left(x\right)={sin}^{2}x=\left(\mathrm{sin }x\right)\left(\mathrm{sin }x\right)$, we use the product rule.

As in Example 4.31, we have

$f^{\prime }\left(x\right)=2f\left(x\right)f^{\prime }\left(x\right)=2\left(\mathrm{sin }x\right)\left(\mathrm{cos }x\right)\text{.}$

We use this information to find the derivative of ${\displaystyle g\left(x\right)={sin}^{3}x=f\left(x\right)\left(\mathrm{sin }x\right)\text{:}}$

${\displaystyle \begin{array}{lll}\hfill g^{\prime }\left(x\right)& =f\left(x\right){\displaystyle \frac{d}{dx}}\sin x+f^{\prime }\left(x\right)\left(\sin x\right)\hfill & \hfill \\ \hfill & =\left({\sin }^{2}x\right)\left(\cos x\right)+\left(2\mathrm{ sin}x\cos x\right)\left(\sin x\right)\hfill & \hfill \\ \hfill & =3{\mathrm{ sin}}^{2}x\cos x\hfill & \hfill \end{array}}$