Trig Derivatives, Math Help and Tutoring

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Introduction

Thus far we have only looked at computing derivatives of polynomial functions (because we derived the power rule from the definition of the derivativeand then took advantage of the linearity of the derivative to handle additions and subtractions of functions) and other associated rules for differentiation (including the product rule, quotient rule, and chain rule). However, there are other important functions for which we need to be able to compute the derivative. Most of what you learn throughout this class comes from applications in science; physics, in particular, requires countless derivatives of many different functions.

I provide the following trigonometric derivatives here without proof for usage on any number of problems. Should you need more description about these relationships, more information about why they’re true (including 2 different levels of proof), or more practice problems, check out my full notes on this section. The following should be memorized:

(1) $\begin{align*} & \diff{}{x}\Big[\sin{x} \Big] = \cos{x} && \diff{}{x}\Big[\cos{x} \Big] = - \sin{x} && \diff{}{x}\Big[\tan{x} \Big] = \sec^2{x} \end{align*}$

(2) $\begin{align*} & \diff{}{x}\Big[\csc{x} \Big] = -\csc{x}\cot{x} && \diff{}{x}\Big[\sec{x} \Big] = \sec{x}\tan{x} && \diff{}{x}\Big[\cot{x} \Big] = -\csc^2{x} \end{align*}$

(3) $\begin{align*} & \diff{}{x}\Big[\sin{x} \Big] = \cos{x} \\ & \diff{}{x}\Big[\cos{x} \Big] = - \sin{x} \\ & \diff{}{x}\Big[\tan{x} \Big] = \sec^2{x} \end{align*}$

(4) $\begin{align*} & \diff{}{x}\Big[\csc{x} \Big] = -\csc{x}\cot{x} \\ & \diff{}{x}\Big[\sec{x} \Big] = \sec{x}\tan{x} \\ & \diff{}{x}\Big[\cot{x} \Big] = -\csc^2{x} \end{align*}$

It should also be noted that the chain rule applies for all compositions of functions (review that section if you need more practice with this idea and how to identify when you have compositions of functions). As a result, if instead of $x$ in the above formulas, you come across an entire function inside of the trig function, you still have to multiply the result by a derivative of what’s “inside” as follows:

$\diff{}{x}\Big[\sin{\big( f(x) \big)} \Big] = \cos{\big( f(x) \big)} \cdot f'(x)$

(5) $\begin{equation*} \diff{}{x} \Big[\cos{\big( f(x) \big)} \Big] = -\sin{\big( f(x) \big)} \cdot f'(x) \end{equation*}$

$\vdots$

etc.

Notice that during the first part of the chain rule, the inner argument inside of the trig function is unchanged: $f(x)$ remains inside. However, the derivative of $f(x)$ multiplies this entire term. Let’s look at some examples!

Trig Derivative Examples

Example 1

Let $f(x) = 5 \cos x$ Find $f'(x)$ .

We appeal to the above relationships in computing this derivative, coupled with the standard techniques we’ve found for calculating derivatives. The constant pulls out, so the derivative is simply:

$f'(x) = -5 \sin x.$

Example 2

Find the derivative of $z = -2\tan x -4x^2$ .

The derivative operator is still linear, so we’re good to simply take the derivative of the pieces and add/subtract the results just like before! The derivative is

$\diff{z}{x} = -2\sec^2 x -8x$

Example 3

Let $h(t) = 4\cos t - 5\sin t - \csc t$ .

Taking the derivatives of the individual terms produces

$h'(t) = -4\sin t -5\cos t + \csc t \cot t$

where the final term because positive because $\diff{}{t} \csc t = -\csc t \cot t$ .

Example 4

Find the derivative of $g(x) = \frac{4}{3}x^3\sec x$ .

This is a product of two functions, so overall, we require the product rule. Don’t worry about taking the derivative of the secant function until it becomes necessary. Applying the product rule gives the following:

$g'(x) = \diff{}{x} \Big[ \frac{4}{3}x^3 \Big] \cdot \sec x + \frac{4}{3}x^3 \cdot \diff{}{x} \Big[\sec x \Big]$

Now it’s clear where and how we need the trig derivative. Continuing by evaluating these individual derivatives produces

$= 4x^2 \sec x + \frac{4}{3}x^3 \sec x \tan x$

Example 5

Compute the derivative of $y = \cos (3x^6)$ .

Now we have a trigonometric function whose argument is more than a simple variable, but rather an entire function: $3x^6$ . This requires usage of the chain rule, which proceeds as follows:

$\diff{y}{x} = -\sin (3x^6) \cdot (18x^5)$

where the final term is the derivative of what’s inside of the original cosine function.

Example 6

Compute the derivative of $h = \csc(\sqrt{t-2})$ .

Using the chain rule, we get the following:

$\diff{h}{t} = -\csc(\sqrt{t-2})\cot(\sqrt{t-2}) \cdot \frac{1}{2}(t-2)^{-1/2} \cdot (1)$

$= \frac{-\csc(\sqrt{t-2})\cot(\sqrt{t-2})}{2\sqrt{t-2}}$

Example 7

Find $\diff{y}{x}$ for $y= (\cot{4x})^8$ .

See if you can get this derivative before continuing on!

At its heart, this problem is actually a power rule problem, not a trigonometric problem. We have a function to a power, which requires the power rule and the chain rule:

$\diff{y}{x} = 8(\cot{4x})^7 \cdot \diff{}{x} \Big[ \cot{4x} \Big]$

and now we can deal with the trigonometric derivative. If you’re ever unsure about finding these derivatives with the chain rule, write them out like this. Now,

$\diff{}{x} \Big[ \cot{4x} \Big] = -\csc{4x}\cot{4x} \cdot (4)$