# 5.4: Derivatives of Trigonometric Functions

The functions sinx and cosx are periodic, with period 2π. You have learned that the derivative of a differentiable function gives the slope of the tangent line at a point. Before proceeding with this lesson, look at the function curves for the two functions, and see if you can identify any points where you know what the derivative will be. For each function, do these sets of points repeat as x increases or decreases? How often? Can you make a general statement about the derivatives of the trigonometric functions?

## Derivatives of Trigonometric Functions

We now want to find an expression for the derivative of each of the six trigonometric functions:

 sin x cos x tan x csc x sec x cot x

We first consider the problem of differentiating sin x, using the definition of the derivative.

$\displaystyle \frac{d}{dx}[sinx]=\lim_{h \to 0} \frac{sin(x+h)−sinx}{h} \nonumber$

Since

$sin(α+β)=sinαcosβ+cosαsinβ \nonumber$

The derivative becomes

$\displaystyle \frac{d}{dx}[sinx]=\lim_{h \to 0} \frac{sinxcosh+cosxsinh−sinx}{h} \nonumber$

$\displaystyle = \lim_{h \to 0} [sinx(\frac{cosh−1}{h})+cosx(\frac{sinh}{h})] \nonumber$

$\displaystyle =−sinx⋅limh→0(1−coshh)+cosx⋅limh→0(sinhh) \nonumber$

$=−sinx⋅(0)+cosx⋅(1) \nonumber$

$=cosx \nonumber$

Let’s look at $\displaystyle \lim_{h\to0}\frac{1−cos(h)}{h} \nonumber$ and $\displaystyle \lim_{x \to 0} \frac{sin(h)}{h}\nonumber$ While there are analytical methods that can be used to evaluate these two limits, let’s look at function graphs and table data. The graphs of the two functions and some calculator table data are shown below. Inspection of these near x=0 appears to show that the limits are:

$\displaystyle \lim_{x \to 0} \frac{cos(h)−1}{h}=0 \nonumber$ and $\displaystyle \lim_{h \to 0} \frac{sin(h)}{h}=1 \nonumber$

The above results can be confirmed analytically.

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 x(rad) -0.001 -0.0001 0 0.0001 0.0001 f(x) -0.0005 -5e-05 Error 5e-05 0.0005 g(x) 0.999998 1 Error 1 0.999998

Therefore,

$\frac{d}{dx}[sinx]=cosx \nonumber$

It will be left as an exercise to prove that $\frac{d}{dx}[cosx]=−sinx$

The derivatives of the six trigonometric functions are shown below.

 $\frac{d}{dx} [sinx]= cosx \nonumber$ $\frac{d}{dx} [cosx]= -sinx \nonumber$ $\frac{d}{dx} [tanx]= sec^2x \nonumber$ $\frac{d}{dx} [cscx]= -cscxcotx \nonumber$ $\frac{d}{dx} [secx]= secxtanx \nonumber$ $\frac{d}{dx} [cotx]= -csc^2x \nonumber$

Keep in mind that the argument x for all the trigonometric functions is measured in radians.

All of the derivatives can be proved by the definition of the derivative, but the reciprocal functions can be found using a simpler method. The proof of $\frac{d}{dx}[tanx]=sec^2x \nonumber$ is as follows:

Since

$tanx= \frac{sinx}{cosx} \nonumber$

then

$\frac{d}{dx}[tanx]=\frac{d}{dx}[\frac{sinx}{cosx}] \nonumber$

Using the quotient rule,

$=\frac{(cosx)(cosx)−(sinx)(−sinx)}{cos^2x} \nonumber$

$=\frac{cos^2x+sin^2x}{cos^2x} \nonumber$

$=\frac{1}{cos2x} \nonumber$

$=sec^2x\nonumber$

## Examples

### Example 1

Earlier, you were asked if there are any repeating points for the derivatives of trigonometric functions and if so, how often they repeat.

First, see if you can identify any points where you know the derivative of sinx and cosx: Each function has two places in the interval 0≤x≤2π where the tangent line has a slope of 0.

For each function, do these sets of points repeat as x increases or decreases? Yes, these 0 slope points do repeat for each function.

How often? The pair of 0 slope points repeats with a period of 2π.

Can you make a general statement about the derivatives of the trigonometric functions? Because the function values repeat every period, the derivative of each function at a specific point should also repeat with period 2π.

### Example 2

Find f′(x) if f(x)=x2cosx+sinx.

Using the product rule and the formulas above, we obtain

$f′(x)=x^2(−sinx)+2xcosx+cosx \nonumber$

$=−x^2sinx+2xcosx+cosx \nonumber$

### Example 3

Find dy/dx if $y=\frac{cosx}{1−tanx} \nonumber$ What is the slope of the tangent line at $x=π3?\nonumber$

Using the quotient rule and the formulas above, we obtain

$\frac{dy}{dx}=\frac{(1−tanx)(−sinx)−(cosx)(−sec^2x)}{(1−tanx)^2} \nonumber$

$=\frac{−sinx+tanxsinx+cosxsec^2x}{(1−tanx)^2} \nonumber$

To calculate the slope of the tangent line, we simply substitute $x=\frac{π}{3}\nonumber$

$\frac{dy}{dx}|_{x=\frac{π}{3}}=\frac{−sin(\frac{π}{3})+tan(π3)sin(\frac{π}{3})+cos(\frac{π}{3})sec^2(\frac{π}{3})}{(1−tan(\frac{π}{3}))^2} \nonumber$

We finally get the slope to be approximately

$\frac{dy}{dx}|_{x=\frac{π}{3}}=4.9 \nonumber$

### Example 4

Find dy/dx if $y=\frac{cotx}{sinx} \nonumber$

$\frac{d}{dx}[\frac{cotx}{sinx}]=\frac{sinx\frac{dy}{dx}[cotx]−cotx\frac{dy}{dx}[sinx]}{sin^2x} \nonumber$

Quotient Rule

$=\frac{sinx⋅[−csc^2x]−cotx⋅[cosx]}{sin^2x} \nonumber$

Trigonometric derivatives

$=\frac{−sinx⋅[csc^2x+cot^2x]}{sin^2x} \nonumber$

factor - sin x

$=−cscx⋅[1+cot^2x+cot^2x] \nonumber$

simplify and use trig identities

$=−cscx⋅[1+2cot^2x] \nonumber$

Therefore,

$\frac{dy}{dx}=−cscx⋅[1+2cot^2x] \nonumber$

## Review

For #1-10, find the derivative y′.

1. $y=xsinx+2 \nonumber$
2. $y=x^2cosx−xtanx−1 \nonumber$
3. $y=sin^2x \nonumber$
4. $y=\frac{sinx−1}{sinx+1} \nonumber$
5. $y=\frac{cosx+sinx}{cosx−sinx} \nonumber$
6. $y=\frac{x^{0.5}}{tanx+2} \nonumber$
7. $y=cscxsinx+x \nonumber$
8. $y=\frac{secx}{cscx} \nonumber$
9. If y = csc x, find y′(π/6).
10. y=x5cos(x)?
11. What is the derivative of x2csc(x)?
12. What is the derivative of csc(x)tan(x)?
13. Use the quotient rule to verify that the derivative of sec(x) is sec(x)tan(x).
14. What is the derivative of cot(π/2−x)?
15. What is the derivative of csc2(x)−cot2(x)?

## Vocabulary

Term Definition
derivative The derivative of a function is the slope of the line tangent to the function at a given point on the graph. Notations for derivative include f′(x), dydx, y′, dfdx and \frac{df(x)}{dx}.