## 17Calculus Derivatives - Trigonometric Functions

This page discusses the derivatives of trig functions. There are no tricks in these derivatives. You just need to learn a few simple formulas.
You do not need to know the chain rule for the first part of this page, we discuss the basic derivatives first. Once you have learned the chain rule, you can come back here to work the practice problems.

Basic Trig Derivatives (no chain rule required)

This first section discuss the basics of trig derivatives and does not require you to know the chain rule. Here are the basic rules. If you work enough practice problems, you won't need to memorize them since you will just know them.

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

The first two, sine and cosine, are pretty straightforward since they are ALMOST inverses of each other. You just need to remember the negative sign on the second one.The rest of them can be derived from the sine and cosine rules using the product rule, quotient rule and basic trigonometric identities. Here is a good video showing this derivation.

### PatrickJMT - Deriving the Derivative Formulas for Tangent, Cotangent, Secant, Cosecant [9min-13secs]

video by PatrickJMT

Notice how the derivatives seem to be in similar pairs. Take a few minutes and look for patterns and similarities as well as differences in all three sets. Take special notice of when a negative sign appears.

Okay, let's take a minute and watch a quick video clip for another perspective on how to remember these derivatives.

### Krista King Math - Derivatives of TRIG FUNCTIONS [7min-49secs]

video by Krista King Math

Here are some practice problems. Calculate the derivative of these functions.

Basic Problems

$$\displaystyle{x^4\tan(x)}$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{x^4\tan(x)}$$

$$\displaystyle{ x^3[x \sec^2(x)+4 \tan(x)] }$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{x^4\tan(x)}$$

Solution

$$\displaystyle{\frac{d}{dx}[ x^4 \tan(x) ] = x^4 \sec^2(x) + 4x^3 \tan(x) = }$$ $$\displaystyle{x^3[x \sec^2(x)+4 \tan(x)]}$$

Note: Not factoring out the $$x^3$$ term, may cost you points. Check with your instructor to see what they expect.

$$\displaystyle{ x^3[x \sec^2(x)+4 \tan(x)] }$$

$$\displaystyle{f(x)=\cos(x)\sin(x)}$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{f(x)=\cos(x)\sin(x)}$$

$$f'(x) = \cos^2(x) - \sin^2(x)$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{f(x)=\cos(x)\sin(x)}$$

Solution

 $$\displaystyle{ f'(x) = \frac{d}{dx}\left[ \cos(x) \sin(x) \right] }$$ $$\displaystyle{ f'(x) = \cos(x) \frac{d}{dx}[ \sin(x)] + \sin(x) \frac{d}{dx}[ \cos(x)] }$$ $$f'(x) = \cos(x)[ \cos(x)] + \sin(x)[-\sin(x)]$$ $$f'(x) = \cos^2(x) - \sin^2(x)$$

$$f'(x) = \cos^2(x) - \sin^2(x)$$

$$\displaystyle{\frac{\sin(x)}{1+\cos(x)}}$$

Problem Statement

Use the quotient rule to calculate the derivative of $$\displaystyle{\frac{\sin(x)}{1+\cos(x)}}$$.

$$\displaystyle{ \frac{d}{dx}\left[ \frac{\sin(x)}{1+\cos(x)} \right] = \frac{1}{1+\cos(x)}}$$

Problem Statement

Use the quotient rule to calculate the derivative of $$\displaystyle{\frac{\sin(x)}{1+\cos(x)}}$$.

Solution

 $$\displaystyle{ \frac{d}{dx}\left[ \frac{\sin(x)}{1+\cos(x)} \right] }$$ $$\displaystyle{ \frac{[1+\cos(x)]d[\sin(x)]/dx - [\sin(x)]d[1+\cos(x)]/dx}{[1+\cos(x)]^2} }$$ $$\displaystyle{ \frac{[1+\cos(x)][\cos(x)] - [\sin(x)][-\sin(x)]}{[1+\cos(x)]^2} }$$ $$\displaystyle{ \frac{\cos(x)+\cos^2(x)+\sin^2(x)}{[1+\cos(x)]^2} }$$ Use the identity $$\sin^2(x) + \cos^2(x) = 1$$. $$\displaystyle{ \frac{1+\cos(x)}{[1+\cos(x)]^2} }$$ $$\displaystyle{ \frac{1}{1+\cos(x)} }$$

$$\displaystyle{ \frac{d}{dx}\left[ \frac{\sin(x)}{1+\cos(x)} \right] = \frac{1}{1+\cos(x)}}$$

$$\displaystyle{y=\frac{\tan(x)}{x^{3/2}+5x}}$$

Problem Statement

Use the quotient rule to calculate the derivative of $$\displaystyle{y=\frac{\tan(x)}{x^{3/2}+5x}}$$

Solution

### 955 video

video by PatrickJMT

Intermediate Problems

$$\displaystyle{\frac{3}{x}\cot(x)}$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{\frac{3}{x}\cot(x)}$$

$$\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] = \frac{-3}{x^2}\left[ x \csc^2(x) + \cot(x) \right] }$$

Problem Statement

Use the product rule to calculate the derivative of $$\displaystyle{\frac{3}{x}\cot(x)}$$

Solution

 $$\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] }$$ $$\displaystyle{ \frac{3}{x}\frac{d}{dx}[\cot(x)] + \cot(x) \frac{d}{dx}[3/x] }$$ $$\displaystyle{ \frac{3}{x} (-\csc^2(x)) + \cot(x) \frac{d}{dx}[3x^{-1}] }$$ $$\displaystyle{ \frac{-3}{x} \csc^2(x) + \cot(x) [3(-1)x^{-2}] }$$ $$\displaystyle{ \frac{-3}{x} \csc^2(x) - \frac{3}{x^2} \cot(x) }$$ $$\displaystyle{ \frac{-3}{x^2}[ x \csc^2(x) + \cot(x) ] }$$

$$\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] = \frac{-3}{x^2}\left[ x \csc^2(x) + \cot(x) \right] }$$

Trig Derivatives Using The Chain Rule

This section requires understanding of the chain rule and the previous section on the basics of trig derivatives. Except for the most basic problems, you will find most problems require the use of the chain rule.
As we discuss on the chain rule page, we think the easiest way to work problems that require the chain rule is to start on the outside and work your way in. Doing it the other direction, requires you to know where to start and what is actually on the inside. Especially when starting out, this is more complicated in our opinion. ( Of course, you need to consult your instructor to find out what they require. )

The best way to work these derivatives is to use substitution, especially while you are still learning. Later, you can do it in your head. In the examples on this page, we will write out the substitution to help you see it.

Let's start out by watching a video. This is a great video explaining the idea of the chain rule when a trig function is involved.

### Krista King Math - Chain rule for derivatives, with trig functions [5min-38secs]

video by Krista King Math

Okay, let's work some examples. If you feel up to it, try to work these before looking at the solutions.

Evaluate $$[\sin(x^2)]'$$

$$[\sin(x^2)]' = 2x \cos(x^2)$$

Problem Statement - Evaluate $$[\sin(x^2)]'$$
Solution - We can tell that the outside function is sine. So, we use the substitution $$u = x^2$$. The reason we chose $$x^2$$ is because we want $$u$$ to be equal to everything inside the sine function.

 $$\displaystyle{ \frac{d}{dx}[\sin(x^2)] }$$ let $$u=x^2$$ $$\displaystyle{ \frac{d}{dx}[\sin(u)] }$$ Apply the chain rule. $$\displaystyle{ \frac{d}{du}[\sin(u)] \frac{d}{dx}[u] }$$ $$\displaystyle{ \frac{d}{du}[\sin(u)] \frac{d}{dx}[x^2] }$$ $$\displaystyle{ \cos(u) (2x) }$$ Reverse the substitution of $$u$$. $$\displaystyle{ 2x \cos(x^2) }$$

Note -
Once we have taken all derivatives, we have $$\cos(u)(2x)$$. We can't leave the answer in this form. Remember, the substitution technique is used only temporarily during computations and none of the temporary variables should appear in the final answer.

$$[\sin(x^2)]' = 2x \cos(x^2)$$

Evaluate $$[\sec(x^2)]'$$

$$[\sec(x^2)]' = 2x \sec(x^2) \tan(x^2)$$

Problem Statement - Evaluate $$[\sec(x^2)]'$$
Solution - The outside function is secant. So, we will again use the substitution $$u = x^2$$.

 $$\displaystyle{ \frac{d}{dx}[\sec(x^2)] }$$ $$\displaystyle{ \frac{d}{dx}[\sec(u)] }$$ $$\displaystyle{ \frac{d}{du}[\sec(u)] \frac{d}{dx}[u] }$$ $$\displaystyle{ \frac{d}{du}[\sec(u)] \frac{d}{dx}[x^2] }$$ $$\sec(u) \tan(u) (2x)$$ $$2x \sec(x^2) \tan(x^2)$$

Note:
There is really nothing new here that wasn't in the first example, except that, at the end, there is more than one place where we need to substitute $$u$$ back in, i.e. $$\sec(u) ~~ \to ~~ \sec(x^2)$$ and $$\tan(u) ~~ \to ~~ \tan(x^2)$$.

$$[\sec(x^2)]' = 2x \sec(x^2) \tan(x^2)$$

This next example uses the product rule and derivative of exponentials. If you have not studied both of these topics, you can skip this example and come back to it once you have. This example also demonstrates how to work a nested chain rule problem.

Evaluate $$[\csc(2xe^{15x})]'$$

$$\displaystyle{ [\csc(2xe^{15x})]' = -2 (15x+1) e^{15x} \csc(2xe^{15x}) \cot(2xe^{15x}) }$$

Problem Statement -
Solution - Okay, as before, we use substitution, letting $$u$$ be equal to everything inside the cosecant term, i.e. $$u=2xe^{15x}$$.

 $$\displaystyle{ \frac{d}{dx}[\csc(2xe^{15x})] }$$ Let $$u=2xe^{15x}$$ $$\displaystyle{ \frac{d}{dx}[\csc(u)] }$$ Apply the chain rule. $$\displaystyle{ \frac{d}{du}[\csc(u)] \frac{d}{dx}[u] }$$ $$\displaystyle{ -\csc(u) \cot(u) \frac{d}{dx}[2xe^{15x}] }$$ Apply the product rule to $$\displaystyle{ \frac{d}{dx}[2xe^{15x}] }$$. $$-\csc(u) \cot(u)$$ $$\displaystyle{ \left[(2x) \frac{d}{dx}[e^{15x}] + e^{15x} \frac{d}{dx}[2x] \right] }$$ Apply the chain rule to $$\displaystyle{ \frac{d}{dx}[e^{15x}] }$$. $$-\csc(u) \cot(u)$$ $$\displaystyle{ \left[(2x) (e^{15x}) \frac{d}{dx}[15x] + e^{15x} (2) \right] }$$ Take remaining derivatives. $$-\csc(u) \cot(u)$$ $$\left[(2x) (e^{15x}) (15) + e^{15x} (2) \right]$$ Reverse substitute for u, factor and simplify. $$-2 (15x+1) e^{15x} \csc(2xe^{15x}) \cot(2xe^{15x})$$

$$\displaystyle{ [\csc(2xe^{15x})]' = -2 (15x+1) e^{15x} \csc(2xe^{15x}) \cot(2xe^{15x}) }$$

Here are some practice problems. Unless otherwise instructed, calculate the derivative of these functions.

Basic Problems

$$7 \sin(x^2+1)$$

Problem Statement

Use the chain rule to calculate the derivative of $$7 \sin(x^2+1)$$

$$(14x) \cos(x^2+1)$$

Problem Statement

Use the chain rule to calculate the derivative of $$7 \sin(x^2+1)$$

Solution

 $$\displaystyle{ \frac{d}{dx}[ 7\sin(x^2+1) ] }$$ Use the constant multiple rule to move the 7 outside the derivative. $$\displaystyle{ 7\frac{d}{dx}[\sin(x^2+1)] }$$ Apply the chain rule. $$\displaystyle{ 7\cos(x^2+1)\frac{d}{dx}[x^2+1] }$$ Take the derivative of the last term. $$7\cos(x^2+1)(2x)$$ Simplify. $$(14x)\cos(x^2+1)$$

A more explicit way to work this is to use substitution, as follows.

 $$\displaystyle{ \frac{d}{dx}[ 7\sin(x^2+1) ] }$$ Let $$u = x^2+1$$ which is the inside term. $$\displaystyle{ 7\frac{d}{du}[\sin(u)] \cdot \frac{d}{dx}[x^2+1] }$$ $$7[\cos(u)](2x)$$ Make sure and replace the u with $$x^2+1$$, so that you don't have any u's in your final answer. And finally, simplify. $$(14x)\cos(x^2+1)$$

If we call the original function, say $$f(x)=7\sin(x^2+1)$$, the first line is $$\displaystyle{ \frac{df}{dx} = \frac{df}{du}\cdot\frac{du}{dx} }$$ using the Chain Rule. This is probably a better way to work the problem than the first solution, especially while you are learning. Once you have the chain rule down, you can easily work it as shown in the first solution.

$$(14x) \cos(x^2+1)$$

$$y=\sin^3(x) \tan(4x)$$

Problem Statement

Use the chain rule to calculate the derivative of $$y=\sin^3(x) \tan(4x)$$

Solution

### 981 video

video by Krista King Math

$$y = (2x^5-3) \sin(7x)$$

Problem Statement

Use the chain rule to calculate the derivative of $$y = (2x^5-3) \sin(7x)$$

$$y' = 7(2x^5-3)\cos(7x) + 10x^4\sin(7x)$$

Problem Statement

Use the chain rule to calculate the derivative of $$y = (2x^5-3) \sin(7x)$$

Solution

 $$\displaystyle{ \frac{dy}{dx} = (2x^5-3) \frac{d}{dx}[\sin(7x)] + \sin(7x)\frac{d}{dx} [2x^5-3] }$$ $$y' = (2x^5-3)\cos(7x) (7) + \sin(7x) (10x^4)$$ $$y' = 7(2x^5-3)\cos(7x) + (10x^4)\sin(7x)$$

$$y' = 7(2x^5-3)\cos(7x) + 10x^4\sin(7x)$$

$$\tan(x^5 + 2x^3 - 12x)$$

Problem Statement

Use the chain rule to calculate the derivative of $$\tan(x^5 + 2x^3 - 12x)$$

$$\displaystyle{ \frac{d}{dx}[\tan(x^5 + 2x^3 - 12x)] = (5x^4 + 6x^2 - 12) \sec^2(x^5 + 2x^3 - 12x) }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\tan(x^5 + 2x^3 - 12x)$$

Solution

Using substitution, the outside function is tangent and the inside function is everything inside the tangent parentheses. So, we let $$u = x^5 + 2x^3 - 12x$$.

 $$\displaystyle{ \frac{d}{dx}[\tan(x^5 + 2x^3 - 12x)] }$$ $$\displaystyle{ \frac{d}{dx}[\tan(u)] }$$ $$\displaystyle{ \frac{d}{du}[\tan(u)] \frac{d}{dx}[u] }$$ $$\displaystyle{ \frac{d}{du}[\tan(u)] \frac{d}{dx}[x^5 + 2x^3 - 12x] }$$ $$\sec^2(u) (5x^4 + 6x^2 - 12)$$ $$(5x^4 + 6x^2 - 12) \sec^2(x^5 + 2x^3 - 12x)$$

$$\displaystyle{ \frac{d}{dx}[\tan(x^5 + 2x^3 - 12x)] = (5x^4 + 6x^2 - 12) \sec^2(x^5 + 2x^3 - 12x) }$$

Intermediate Problems

$$y = (1+\cos^2(7x))^3$$

Problem Statement

Use the chain rule to calculate the derivative of $$y = (1+\cos^2(7x))^3$$

Solution

### 982 video

video by Krista King Math

$$y = \cot^4(x/2)$$

Problem Statement

Use the chain rule to calculate the derivative of $$y = \cot^4(x/2)$$

$$y' = -2\cot^3(x/2)\csc^2(x/2)$$

Problem Statement

Use the chain rule to calculate the derivative of $$y = \cot^4(x/2)$$

Solution

### 1910 video

video by Krista King Math

$$y' = -2\cot^3(x/2)\csc^2(x/2)$$

Calculate the third derivative of $$f(x)=\tan(3x)$$.

Problem Statement

Calculate the third derivative of $$f(x)=\tan(3x)$$.

$$54\sec^2(3x)\left[ 2\tan^2(3x)+\sec^2(3x)\right]$$

Problem Statement

Calculate the third derivative of $$f(x)=\tan(3x)$$.

Solution

$$f'(x)=3\sec^2(3x)$$
$$f''(x)=18\sec^2(3x)\tan(3x)$$

### 2085 video

video by PatrickJMT

$$54\sec^2(3x)\left[ 2\tan^2(3x)+\sec^2(3x)\right]$$

$$\displaystyle{ y=\left[ \tan \left( \sin \left( \sqrt{x^2+8x}\right) \right) \right]^5 }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ y=\left[ \tan \left( \sin \left( \sqrt{x^2+8x}\right) \right) \right]^5 }$$

Solution

### 988 video

video by PatrickJMT

$$\displaystyle{ y=x \sin(1/x) + \sqrt{(1-3x)^2 + x^5} }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ y=x \sin(1/x) + \sqrt{(1-3x)^2 + x^5} }$$

Solution

In this video, his second term incorrectly contains t instead of x.

### 990 video

video by PatrickJMT

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