New Zealand
Level 8 - NCEA Level 3

# Basic derivatives of tan

Lesson

We understand the tangent function in right-angled triangles to be the ratio of sides $\frac{\text{opposite}}{\text{adjacent}}$oppositeadjacent with respect to one of the non right-angles in the triangle. More generally, we use the unit circle definitions of the trigonometric functions in order to give meaning to functions of any angle.  The tangent function is defined in terms of the sine and cosine functions.

$\tan x\equiv\frac{\sin x}{\cos x}$tanxsinxcosx

The graph below shows the tangent function between $-\pi$π and $\pi$π

Observe that the function has asymptotes at $\frac{\pi}{2}\pm n\pi$π2±nπ, the points where the cosine function in the denominator in the definition is zero.

From the graph, it can be seen that the gradient of the tangent function is always positive and it has a minimum value of $1$1 at $x=0\pm n\pi$x=0±nπ. The gradient has no maximum.

We deduce the tangent function explicitly by differentiating $f(x)=\tan x=\frac{\sin x}{\cos x}$f(x)=tanx=sinxcosx. It is convenient to use the quotient rule for this.

We find, $f'\left(x\right)=\frac{\cos x\times\cos x-\left(-\sin x\times\sin x\right)}{\cos^2x}=\frac{\cos^2x+\sin^2x}{\cos^2x}=\frac{1}{\cos^2x}=\sec^2x$f(x)=cosx×cosx(sinx×sinx)cos2x=cos2x+sin2xcos2x=1cos2x=sec2x

In Leibniz's notation, if $y=\tan x$y=tanx, then $\frac{\mathrm{d}y}{\mathrm{d}x}=\sec^2x$dydx=sec2x.

In the diagram below, the graph of $\sec x^2$secx2 (in red) has been superimposed on the tangent graph. It confirms the observations made above about the gradient of the tangent function.

Expressions of the type $a\tan\left(bx+c\right)+d$atan(bx+c)+d are differentiated by the usual rules, so that $\frac{\mathrm{d}}{\mathrm{d}x}\left(a\tan\left(bx+c\right)+d\right)=ab\sec^2\left(bx+c\right)$ddx(atan(bx+c)+d)=absec2(bx+c).

It is essential that the argument of the tangent function be expressed in radian measure. This requirement comes from the way in which the derivative of the sine function was obtained.

##### Example 1

Differentiate $f\left(x\right)=4\tan x+4$f(x)=4tanx+4.

##### Example 2

Differentiate $y=\tan3x$y=tan3x.

##### Example 3

At what values of $x$x does $\tan2x$tan2x have a gradient of $2$2?

We need to find out where the derivative is equal to $2$2. That is, we must solve $2\sec^22x=2$2sec22x=2. So, $\sec^22x=1$sec22x=1. From the graph above, it is clear that $\sec^22x=1$sec22x=1 at $2x=0,\pm\pi,\pm2\pi,...$2x=0,±π,±2π,.... Hence, the required solutions are $x=0,\pm\frac{\pi}{2},\pm\pi,...$x=0,±π2,±π,....

##### Example 4

Consider the graph of $y=\tan x$y=tanx.

1. Which of the following best describes the graph of $y=\tan x$y=tanx?

The graph increases and decreases periodically.

A

It is constantly decreasing.

B

It is constantly increasing.

C

The graph increases and decreases periodically.

A

It is constantly decreasing.

B

It is constantly increasing.

C
2. Which of the following best describes the nature of the gradient of the curve?

Select all the correct options.

A

The gradient to the curve is never $0$0.

B

The gradient function has the same period as the curve itself.

C

The gradient increases more and more rapidly as the curve approaches the asymptotes.

D

E

The gradient to the curve is $0$0 every $\pi$π radians.

F

A

The gradient to the curve is never $0$0.

B

The gradient function has the same period as the curve itself.

C

The gradient increases more and more rapidly as the curve approaches the asymptotes.

D

E

The gradient to the curve is $0$0 every $\pi$π radians.

F
3. The tangent lines at the intercepts of the curve have been graphed as well.

Using the graph, write down the gradient to the curve at $x=0$x=0, $\pi$π, $2\pi$2π, $3\pi$3π, $\text{. . .}$. . .

Gradient $=$= $\editable{}$.

4. The gradient function of $y=\tan x$y=tanx is $y'$y. Which of the following is the correct graph of $y'$y for each value of $x$x?

A

B

C

D

A

B

C

D
5. Which of the following is the equation of the gradient function $y'$y?

$y'=\sec^2\left(x\right)$y=sec2(x)

A

$y'=\sec x$y=secx

B

$y'=\csc^2\left(x\right)$y=csc2(x)

C

$y'=\csc x$y=cscx

D

$y'=\sec^2\left(x\right)$y=sec2(x)

A

$y'=\sec x$y=secx

B

$y'=\csc^2\left(x\right)$y=csc2(x)

C

$y'=\csc x$y=cscx

D

##### Example 5

Differentiate $\tan\left(x^\circ\right)$tan(x°).

Solution:

First, the argument $x^\circ$x° needs to be converted to circular measure.

To convert a degrees measure to radians, we can multiply it by $\frac{\pi}{180}$π180. This means $x^\circ$x° is equal to $\frac{x\pi}{180}$xπ180 radians.

Now we can look to differentiate $\tan\left(\frac{x\pi}{180}\right)$tan(xπ180).

Using the chain rule, we get $\frac{d}{dx}\tan\left(\frac{x\pi}{180}\right)=\frac{\pi}{180}\sec^2\left(\frac{x\pi}{180}\right)$ddxtan(xπ180)=π180sec2(xπ180).

### Outcomes

#### M8-11

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

#### 91578

Apply differentiation methods in solving problems