NZ Level 8 (NZC) Level 3 (NCEA) [In development] Differentiating Various Trig Functions (tan)
Lesson

The essential information about the derivative of the tangent function is found in another chapter.

In this chapter, we see how the procedures for differentiating sums, composite functions, products and quotients apply to the trigonometric functions and particularly to functions involving the tangent function.

## Sums

If $h(x)=f(x)+g(x)$h(x)=f(x)+g(x), then, from first principles,

 $h'(x)$h′(x) $=$= $\lim_{\delta\rightarrow0}\frac{f(x+\delta)+g(x+\delta)-\left(f(x)+g(x)\right)}{\delta}$limδ→0f(x+δ)+g(x+δ)−(f(x)+g(x))δ​ $=$= $\lim_{\delta\rightarrow0}\left[\frac{f(x+\delta)-f(x)}{\delta}+\frac{g(x+\delta)-g(x)}{\delta}\right]$limδ→0[f(x+δ)−f(x)δ​+g(x+δ)−g(x)δ​] $=$= $\lim_{\delta\rightarrow0}\frac{f(x+\delta)-f(x)}{\delta}+\lim_{\delta\rightarrow0}\frac{g(x+\delta)-g(x)}{\delta}$limδ→0f(x+δ)−f(x)δ​+limδ→0g(x+δ)−g(x)δ​

and we see that $h'(x)=f'(x)+g'(x)$h(x)=f(x)+g(x), so long as $f$fand $g$g are differentiable functions.

##### Example 1

If $h(x)=\sin x+\tan x$h(x)=sinx+tanx, find $h'(x)$h(x).

We differentiate $\sin x$sinx and $\tan x$tanx separately. The derivatives are $\cos x$cosx and $\sec^2x$sec2x respectively. Therefore the derivative of the sum is the sum of the derivatives: $h'(x)=\cos x+\sec^2x$h(x)=cosx+sec2x.

## Composite Functions

A composite function $h(x)$h(x), given by $h(x)=g\left(f(x)\right)$h(x)=g(f(x)), can be notated $h(x)=(g\circ f)(x)$h(x)=(gf)(x). In evaluating such a function, the 'inside' function $f$f is applied first and then the 'outside' function $g$g is applied to the result.

Note that, for a composite function to make sense, the domain of $f$f has to be chosen so that the  range of $f$f falls withing the domain of $g$g.

If $h(x)=\left(f\circ g\right)(x)$h(x)=(fg)(x), where $f$f and $g$g are differentiable functions with suitable domains and ranges, then, from first principles,

 $h'(x)$h′(x) $=$= $\lim_{\delta\rightarrow0}\frac{f\left(g(x+\delta\right)-f\left(g(x)\right)}{\delta}$limδ→0f(g(x+δ)−f(g(x))δ​

Recognising that $g(x+\delta)-g(x)$g(x+δ)g(x) is the increment in $g(x)$g(x) as $x\rightarrow x+\delta$xx+δ, we multiply by $1=\frac{g(x+\delta)-g(x)}{g(x+\delta)-g(x)}$1=g(x+δ)g(x)g(x+δ)g(x). Thus,

 $h'(x)$h′(x) $=$= $\lim_{\delta\rightarrow0}\frac{f\left(g(x+\delta\right)-f\left(g(x)\right)}{\delta}.\frac{g(x+\delta)-g(x)}{g(x+\delta)-g(x)}$limδ→0f(g(x+δ)−f(g(x))δ​.g(x+δ)−g(x)g(x+δ)−g(x)​ $=$= $\lim_{\delta\rightarrow0}\frac{f\left(g(x+\delta\right)-f\left(g(x)\right)}{g(x+\delta)-g(x)}.\lim_{\delta\rightarrow0}\frac{g(x+\delta)-g(x)}{\delta}$limδ→0f(g(x+δ)−f(g(x))g(x+δ)−g(x)​.limδ→0g(x+δ)−g(x)δ​ $=$= $f'\left(g(x)\right).g'(x)$f′(g(x)).g′(x)

This is the 'chain rule', which has been introduced elsewhere. In essence, it says that we differentiate the 'outside' function with respect to the 'inner' function and multiply by the derivative of the 'inner' function with respect to the variable. In Leibniz's notation, this is $\frac{\mathrm{d}h}{\mathrm{d}x}=\frac{\mathrm{d}h}{\mathrm{d}g}\frac{\mathrm{d}g}{\mathrm{d}x}$dhdx=dhdgdgdx.

##### Example 2

Differentiate $y=\tan^2x$y=tan2x.

We differentiate the squaring function with respect to $\tan x$tanx and we differentiate $\tan x$tanx with respect to $x$x. Then $y'=2\tan x.\sec^2x$y=2tanx.sec2x.

## Products

If $f$f and $g$g are differentiable functions, then, from first principles, the product $h=fg$h=fg has derivative

 $h'(x)$h′(x) $=$= $\lim_{\delta\rightarrow0}\frac{f(x+\delta)g(x+\delta)-f(x)g(x)}{\delta}$limδ→0f(x+δ)g(x+δ)−f(x)g(x)δ​ $=$= $\lim_{\delta\rightarrow0}\frac{f(x+\delta).g(x+\delta)-f(x).g(x+\delta)+f(x).g(x+\delta)-f(x)g(x)}{\delta}$limδ→0f(x+δ).g(x+δ)−f(x).g(x+δ)+f(x).g(x+δ)−f(x)g(x)δ​ $=$= $\lim_{\delta\rightarrow0}\frac{g(x)\left(f(x+\delta)-f(x)\right)+f(x)\left(g(x+\delta)-g(x)\right)}{\delta}$limδ→0g(x)(f(x+δ)−f(x))+f(x)(g(x+δ)−g(x))δ​ $=$= $g(x).\lim_{\delta\rightarrow0}\frac{\left(f(x+\delta)-f(x)\right)}{\delta}+f(x).\lim_{\delta\rightarrow0}\frac{\left(g(x+\delta)-g(x)\right)}{\delta}$g(x).limδ→0(f(x+δ)−f(x))δ​+f(x).limδ→0(g(x+δ)−g(x))δ​ $=$= $f'(x).g(x)+f(x).g'(x)$f′(x).g(x)+f(x).g′(x)

This is the product rule.

##### Example 3

Differentiate the product $h(x)=\sin x.\tan x$h(x)=sinx.tanx.

The derivative of $\sin x$sinx is $\cos x$cosx and the derivative of $\tan x$tanx is $\sec^2x$sec2x. So, $h'(x)=\cos x.\tan x+\sin x.\sec^2x$h(x)=cosx.tanx+sinx.sec2x.

This can be simplified slightly to $\sin x\left(1+\sec^2x\right)$sinx(1+sec2x) or $\sin x\left(2+\tan^2x\right)$sinx(2+tan2x).

## Quotients

We can obtain the quotient rule from the product and composite function rules.

For differentiable functions $f$f and $g$g with appropriate domains, the quotient $h(x)=\frac{f(x)}{g(x)}$h(x)=f(x)g(x) can be written as a product: $h(x)=f(x)\left(g(x)\right)^{-1}$h(x)=f(x)(g(x))1.

Now, $f(x)$f(x) has derivative $f'(x)$f(x) and $\left(g(x)\right)^{-1}$(g(x))1 has derivative $(-1)\left(g(x)\right)^{-2}.g'(x)$(1)(g(x))2.g(x). So, the product has derivative $f'(x).\left(g(x)\right)^{-1}+f(x).(-1)\left(g(x)\right)^{-2}.g'(x)$f(x).(g(x))1+f(x).(1)(g(x))2.g(x).

This is written more simply as: $h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{\left(g(x)\right)^2}$h(x)=f(x)g(x)f(x)g(x)(g(x))2

##### Example 4

Differentiate $\cot x$cotx.

This can be done by writing $\cot x$cotx as $\frac{1}{\tan x}$1tanx. Then, using the quotient rule, the derivative must be $\frac{0\times\tan x-1\times\sec^2x}{\tan^2x}$0×tanx1×sec2xtan2x. This simplifies to $\frac{-1}{\sin^2x}=-\csc^2x$1sin2x=csc2x.

We could also write $\cot x$cotx as $\frac{\cos x}{\sin x}$cosxsinx and again differentiate using the quotient rule.

#### More Examples

##### Question 1

Differentiate $y=x\tan4x$y=xtan4x.

##### Question 2

Differentiate $y=\tan\left(\sin x\right)$y=tan(sinx).

##### Question 3

Differentiate $y=\sqrt{\tan6x}$y=tan6x.

### 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