From the graph, it is clear that the gradient of the sine function also varies periodically - from a maximum of $1$1 at $x=0$x=0, through zero at $x=\frac{\pi}{2}$x=π2, to a minimum of $-1$−1 at $x=\pi$x=π, back to zero at $x=\frac{3\pi}{2}$x=3π2 and then returning to $1$1 at $x=2\pi$x=2π. This cycle is then repeated.

The following graph shows the sine function and the cosine function.

The graphs strongly suggest that if $f\left(x\right)=\sin x$f(x)=sinx, then the derivative is $f'\left(x\right)=\cos x$f′(x)=cosx.

We can undertake a similar process for $f\left(x\right)=\cos x$f(x)=cosx. The graph below shows $f\left(x\right)=\cos x$f(x)=cosx and $g\left(x\right)=-\sin\left(x\right)$g(x)=−sin(x). The gradient of $f\left(x\right)=\cos x$f(x)=cosx at $x=0,\pi$x=0,π is equivalent to the function value of $f\left(x\right)=-\sin x$f(x)=−sinx at these points. Observation of these graphs suggest that if $f\left(x\right)=\cos x$f(x)=cosx, then $f'\left(x\right)=-\sin\left(x\right)$f′(x)=−sin(x) .

Approximating the derivatives using limits

Another way to obtain the derivative of $\sin x$sinx is to use first principles as follows:

$f'\left(x\right)=\lim_{h\rightarrow0}\frac{\sin\left(x+h\right)-\sin x}{h}$f′(x)=limh→0sin(x+h)−sinxh

As we have not introduced the trigonometric identities required to solve this limit problem, we will investigate the limit equation above using selected values of $x$x and an appropriately small value of $h$h. and compare it to values of $\cos x$cosx. Note that values of $x$x and $h$h will be in radians.

$x$`x`	$h$`h`	$f'\left(x\right)=\lim_{h\rightarrow0}\frac{\sin\left(x+h\right)-\sin x}{h}$`f`′(`x`)=lim`h`→0`sin`(`x`+`h`)−`sinxh`	$\cos x$`cosx`
$0.5$0.5	$0.0001$0.0001	$0.87756$0.87756	$087758$087758
$1$1	$0.0001$0.0001	$0.54026$0.54026	$0.54030$0.54030
$1.5$1.5	$0.0001$0.0001	$0.07069$0.07069	$0.07074$0.07074
$2$2	$0.0001$0.0001	$-0.41619$−0.41619	$-0.41615$−0.41615
$2.5$2.5	$0.0001$0.0001	$-0.80117$−0.80117	$-0.80114$−0.80114
$3$3	$0.0001$0.0001	$-0.99000$−0.99000	$-0.98999$−0.98999

It is clear from the table above that the limit equation for the derivative of $f\left(x\right)=\sin x$f(x)=sinx closely approximates $\cos x$cosx.

It does indeed turn out that for $f\left(x\right)=\sin x$f(x)=sinx, the derivative is:

$f'\left(x\right)$f′(x)

$=$=

$\cos x$cosx

A similar investigation for $f\left(x\right)=\cos x$f(x)=cosx will confirm its derivative as:

$f'\left(x\right)$f′(x)

$=$=

$-\sin x$−sinx

The two methods of approximation mentioned above aim to provide an insight, but not a formal proof into the nature of the derivatives of the sine and cosine functions. These rules for the derivatives of the basic sine and cosine functions can be summarised as follows:

Derivatives of $\sin x$sinx and $\cos x$cosx

For $f\left(x\right)=\sin x$f(x)=sinx, $f'\left(x\right)=\cos x$f′(x)=cosx

For $f\left(x\right)=\cos x$f(x)=cosx, $f'\left(x\right)=-\sin x$f′(x)=−sinx

Practice question

Question 1

Consider the graphs of $y=\cos x$y=cosx and its derivative $y'=-\sin x$y′=−sinx below. A number of points have been labelled on the graph of $y'=-\sin x$y′=−sinx.

Which point on the gradient function corresponds to where the graph of $y=\cos x$y=cosx is increasing most rapidly?
$A$A
A
$B$B
B
$C$C
C
$D$D
D
$E$E
E
Which point on the gradient function corresponds to where the graph of $y=\cos x$y=cosx is decreasing most rapidly?
$A$A
A
$B$B
B
$C$C
C
$D$D
D
$E$E
E
Which points on the gradient function corresponds to where the graph of $y=\cos x$y=cosx is stationary?

Select all that apply.
$A$A
A
$B$B
B
$C$C
C
$D$D
D
$E$E
E

The derivatives of $\sin\left(f\left(x\right)\right)$`sin`(`f`(`x`)) and $\cos\left(f\left(x\right)\right)$`cos`(`f`(`x`))

We can develop rules for the differentiation of functions in the form $\sin\left(f\left(x\right)\right)$sin(f(x)) and $\cos\left(f\left(x\right)\right)$cos(f(x)) using the chain rule.

Consider functions of the form:

$y$y

$=$=

$\sin\left(f\left(x\right)\right)$sin(f(x))

Letting $u=f\left(x\right)$u=f(x), we have:

$y$`y`	$=$=	$\sin u$`sinu`
$\frac{dy}{du}$`dydu`	$=$=	$\cos u$`cosu`
$\frac{du}{dx}$`dudx`	$=$=	$f'\left(x\right)$`f`′(`x`)

Using the chain rule:

$\frac{dy}{dx}$`dydx`	$=$=	$\frac{dy}{du}\times\frac{du}{dx}$`dydu`×`dudx`
	$=$=	$\cos u\times u'$`cosu`×`u`′
	$=$=	$f'\left(x\right)\cos\left(f\left(x\right)\right)$`f`′(`x`)`cos`(`f`(`x`))

Similarly for functions of the form:

$y$y

$=$=

$\cos\left(f\left(x\right)\right)$cos(f(x))

Letting $u=f\left(x\right)$u=f(x), we have:

$y$`y`	$=$=	$\cos u$`cosu`
$\frac{dy}{du}$`dydu`	$=$=	$-\sin u$−`sinu`
$\frac{du}{dx}$`dudx`	$=$=	$f'\left(x\right)$`f`′(`x`)

Using the chain rule:

$\frac{dy}{dx}$`dydx`	$=$=	$\frac{dy}{du}\times\frac{du}{dx}$`dydu`×`dudx`
	$=$=	$-\sin u\times u'$−`sinu`×`u`′
	$=$=	$-f'\left(x\right)\sin\left(f\left(x\right)\right)$−`f`′(`x`)`sin`(`f`(`x`))

Differentiating $y=\sin\left(f\left(x\right)\right)$y=sin(f(x)) and $y=\cos\left(f\left(x\right)\right)$y=cos(f(x))

$\frac{d}{dx}\sin\left(f\left(x\right)\right)$`ddx``sin`(`f`(`x`))	$=$=	$f'\left(x\right)\cos\left(f\left(x\right)\right)$`f`′(`x`)`cos`(`f`(`x`))
$\frac{d}{dx}\cos\left(f\left(x\right)\right)$`ddx``cos`(`f`(`x`))	$=$=	$-f'\left(x\right)\sin\left(f\left(x\right)\right)$−`f`′(`x`)`sin`(`f`(`x`))

Practice questions

Question 2

Differentiate $y=8\sin\left(\frac{x}{4}\right)$y=8sin(x4).

Write each line of working as an equation.

Question 3

Differentiate $y=\cos\left(3x^2\right)$y=cos(3x2).

The derivative of the tangent function $y=\tan x$`y`=`tanx`

The tangent function is defined, 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.

$\tan x=\frac{\sin x}{\cos x}$tanx=sinxcosx, for any $x$x not an odd multiple of $\frac{\pi}{2}$π2.

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

Observe that the function has asymptotes at $\pm\frac{\pi}{2}$±π2 which are repeated every $\pi$π radians in both directions. These asymptotes correspond to the points where the cosine function in the denominator of the identity $\tan x=\frac{\sin x}{\cos x}$tanx=sinxcosx is equal to 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 value.

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 as follows:

$f'(x)$`f`′(`x`)	$=$=	$\frac{\cos x\times\cos x-\left(-\sin x\times\sin x\right)}{\cos^2x}$`cosx`×`cosx`−(−`sinx`×`sinx`)`cos`2`x`
	$=$=	$\frac{\cos^2x+\sin^2x}{\cos^2x}$`cos`2`x`+`sin`2`xcos`2`x`

Using the Pythagorean trig identity $\sin^2x+\cos^2x=1$sin2x+cos2x=1:

$f'(x)$`f`′(`x`)	$=$=	$\frac{1}{\cos^2x}$1`cos`2`x`
	$=$=	$\sec^2x$`sec`2`x`

In the diagram below, the graph of $\sec\left(x^2\right)$sec(x2) (in red) has been superimposed on the tangent graph. It confirms the observations made above about the gradient of the tangent function.

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.

Derivative of $\tan x$tanx

For $f\left(x\right)=\tan x$f(x)=tanx, the derivative is given by

$f'\left(x\right)=\sec^2x$f′(x)=sec2x, where $x$x is in radians

The derivative of $\tan\left(f\left(x\right)\right)$`tan`(`f`(`x`))

The derivative of $\tan\left(f\left(x\right)\right)$tan(f(x)) can also be obtained using the chain rule as follows:

Consider functions of the form:

$y$y

$=$=

$\tan\left(f\left(x\right)\right)$tan(f(x))

Letting $u=f\left(x\right)$u=f(x), we have:

$y$`y`	$=$=	$\tan u$`tanu`
$\frac{dy}{du}$`dydu`	$=$=	$\sec^2u$`sec`2`u`
$\frac{du}{dx}$`dudx`	$=$=	$f'\left(x\right)$`f`′(`x`)

Using the chain rule:

$\frac{dy}{dx}$`dydx`	$=$=	$\frac{dy}{du}\times\frac{du}{dx}$`dydu`×`dudx`
	$=$=	$\sec^2u\times u'$`sec`2`u`×`u`′
	$=$=	$f'\left(x\right)\sec^2\left(f\left(x\right)\right)$`f`′(`x`)`sec`2(`f`(`x`))

Derivative of $\tan\left(f\left(x\right)\right)$tan(f(x))

$\frac{d}{dx}\tan\left(f\left(x\right)\right)=f'(x)\sec^2\left(f\left(x\right)\right)$ddxtan(f(x))=f′(x)sec2(f(x)), where $x$x is in radians

Worked example

Example 1

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

Think: First, the argument $x^\circ$x° needs to be converted to radians. 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.

Do: 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).