11.18 Differentiating trigonometric functions
The derivatives of $\sin x$sinx and $\cos x$cosx
Graphical approximation of the derivatives
The sine and cosine functions are periodic, meaning that those functions go through cycles, or repeat.
The period of both the basic sine and cosine functions, $\sin x$sinx and $\cos x$cosx, is $2\pi$2π.
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)=limh→0sin(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:
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
EWhich 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
EWhich 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)) |
| $\frac{d}{dx}\sin\left(f\left(x\right)\right)$ddxsin(f(x)) | $=$= | $f'\left(x\right)\cos\left(f\left(x\right)\right)$f′(x)cos(f(x)) |
| $\frac{d}{dx}\cos\left(f\left(x\right)\right)$ddxcos(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).
Differentiation of harder trigonometric functions
When faced with finding the derivative of a more complicated trigonometric function, we can utilise the product, quotient and other rules to arrive at a solution.
Practice questions
Question 4
Find the derivative of $x^{\frac{7}{4}}\cos2x$x74cos2x.