From the investigation we found and proved the following rules for differentiating trigonometric functions:
If $y=\sin x$y=sinx, then $\frac{dy}{dx}=\cos x$dydx=cosx
and
if $y=\cos x$y=cosx, then $\frac{dy}{dx}=-\sin x$dydx=−sinx.
The derivative 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)) |
Setting $u=f\left(x\right)$u=f(x), we have $y=\sin\left(u\right)$y=sin(u). We then observe that:
$\frac{du}{dx}=f'\left(x\right)$dudx=f′(x) and $\frac{dy}{du}=\cos u$dydu=cosu
Using the chain rule:
| $\frac{dy}{dx}$dydx | $=$= | $\frac{dy}{du}\times\frac{du}{dx}$dydu×dudx |
|
| $=$= | $\cos u\times f'\left(x\right)$cosu×f′(x) |
|
|
| $=$= | $f'\left(x\right)\cos\left(f\left(x\right)\right)$f′(x)cos(f(x)) |
Substitute $u=f\left(x\right)$u=f(x) |
Similarly we can apply the chain rule to $y=\cos\left(f\left(x\right)\right)$y=cos(f(x)) to find its derivative. Try this yourself and confirm the rule given in the box below.
Worked examples
Example 1
Find the derivative of $y=\cos\left(5x\right)$y=cos(5x).
Think: We can see the function is of the form $y=\cos\left(f\left(x\right)\right)$y=cos(f(x)), with $f\left(x\right)=5x$f(x)=5x. Therefore we need to use the chain rule to differentiate.
Do:
| $\frac{dy}{dx}$dydx | $=$= | $-f'\left(x\right)\sin\left(f\left(x\right)\right)$−f′(x)sin(f(x)) |
Multiply the derivative of the inside function by the outside function |
| $=$= | $-5\sin\left(5x\right)$−5sin(5x) |
|
Reflect: We can form a simplified version of the chain rule for the special case where $y=\sin\left(kx\right)$y=sin(kx) or $y=\cos\left(kx\right)$y=cos(kx), where $k$k is a constant. Since the derivative of the inside function is simply $k$k, we obtain the rules:
If $y=\sin\left(kx\right)$y=sin(kx), then $\frac{dy}{dx}=k\cos\left(kx\right)$dydx=kcos(kx) and if $y=\cos\left(kx\right)$y=cos(kx), then $\frac{dy}{dx}=-k\sin\left(kx\right)$dydx=−ksin(kx).
Example 2
Find the derivative of $y=10\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)$y=10sin(x2+π4).
Think: We can see the function is of the form $y=10\sin\left(f\left(x\right)\right)$y=10sin(f(x)), with $f\left(x\right)=\frac{x}{2}+\frac{\pi}{4}$f(x)=x2+π4. Therefore we need to use the chain rule to differentiate. The constant out the front will simply multiply the derivative and $f\left(x\right)$f(x) is a linear function with derivative $f'\left(x\right)=\frac{1}{2}$f′(x)=12.
Do:
| $\frac{dy}{dx}$dydx | $=$= | $10f'\left(x\right)\cos\left(f\left(x\right)\right)$10f′(x)cos(f(x)) |
Multiply the derivative of the inside function by the outside function |
| $=$= | $10\times\frac{1}{2}\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)$10×12cos(x2+π4) |
|
|
| $=$= | $5\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)$5cos(x2+π4) |
Practice questions
Question 1
Differentiate $y=3x+\sin5x$y=3x+sin5x.
Question 2
Differentiate $y=\cos\left(3x^2\right)$y=cos(3x2).
The derivative of the tangent function $y=\tan x$y=tanx
Recall the tangent function can be defined in terms of the sine and cosine functions:
$\tan x=\frac{\sin x}{\cos x}$tanx=sinxcosx
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 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 can deduce the tangent function explicitly by differentiating $f\left(x\right)=\tan x=\frac{\sin x}{\cos x}$f(x)=tanx=sinxcosx using the quotient rule for this as follows:
Do:
| Let | $u=\sin x$u=sinx | then | $u'=\cos x$u′=cosx |
| and | $v=\cos x$v=cosx | then |
$v'=-\sin x$v′=−sinx |
| $f'\left(x\right)$f′(x) | $=$= | $\frac{vu'-uv'}{v^2}$vu′−uv′v2 |
|
| $=$= | $\frac{\cos x\times\cos x-\left(-\sin x\times\sin x\right)}{\cos^2x}$cosx×cosx−(−sinx×sinx)cos2x |
|
|
| $=$= | $\frac{\cos^2x+\sin^2x}{\cos^2x}$cos2x+sin2xcos2x |
|
|
| $=$= | $\frac{1}{\cos^2x}$1cos2x |
Using the identity $\sin^2x+\cos^2x=1$sin2x+cos2x=1 |
While sine, cosine and tangent are the most widely used trigonometric functions, there are several more that are in common use including secant, cosecant and cotangent. These functions can be abbreviated as sec, cosec and cot, and defined in terms of our common functions as follows:
$\sec\theta=\frac{1}{\cos\theta}$secθ=1cosθ, $\operatorname{cosec}\theta=\frac{1}{\sin\theta}$cosecθ=1sinθ and $\cot\theta=\frac{1}{\tan\theta}$cotθ=1tanθ
The derivative of $f\left(x\right)=\tan x$f(x)=tanx, given by $f'\left(x\right)=\frac{1}{\cos^2x}$f′(x)=1cos2x is often written as $f'\left(x\right)=\sec^2x$f′(x)=sec2x.
In the diagram below, the graph of $y=\sec^2x$y=sec2x (in red) has been superimposed on the tangent graph. It confirms the observations made above about the gradient of the tangent function.
If $y=\tan x$y=tanx, then $\frac{dy}{dx}=\frac{1}{\cos^2x}=\sec^2x$dydx=1cos2x=sec2x
and
if $y=\tan\left(f\left(x\right)\right)$y=tan(f(x)), then $\frac{dy}{dx}=\frac{f'\left(x\right)}{\cos^2\left(f\left(x\right)\right)}=f'\left(x\right)\sec^2\left(f\left(x\right)\right)$dydx=f′(x)cos2(f(x))=f′(x)sec2(f(x)).
Worked example
Example 3
Differentiate $y=3\tan\left(\sqrt{x}\right)$y=3tan(√x).
Think: We can see the function is of the form $y=\tan\left(f\left(x\right)\right)$y=tan(f(x)), with $f\left(x\right)=x^{\frac{1}{2}}$f(x)=x12. Therefore we need to use the chain rule to differentiate.
Do:
| $\frac{dy}{dx}$dydx | $=$= | $3\times f'\left(x\right)\sec^2\left(f\left(x\right)\right)$3×f′(x)sec2(f(x)) |
Take the constant out, and multiply the derivative of the inside function by the outside function |
| $=$= | $3\times\frac{1}{2}x^{\frac{-1}{2}}\sec^2\left(x^{\frac{1}{2}}\right)$3×12x−12sec2(x12) |
|
|
| $=$= | $\frac{3\sec^2\left(\sqrt{x}\right)}{2\sqrt{x}}$3sec2(√x)2√x |
Practice questions
Question 3
Differentiate $y=-4\tan5x$y=−4tan5x.
Further differentiation of trigonometric functions
Just as with exponential functions we can be expected to differentiate more complex trigonometric functions with the chain rule, the product rule, the quotient rule or combinations of these. Let's look at some examples.
Worked example
Example 4
Find the derivative of $y=\cos^2\left(3x\right)$y=cos2(3x)
Think: It may be more useful to think of this function written in the form $y=\left[\cos\left(3x\right)\right]^2$y=[cos(3x)]2. It is now clearer we have a function in the form $y=\left[f\left(x\right)\right]^n$y=[f(x)]n and we can proceed in using the chain rule to differentiate.
Do:
| $\frac{dy}{dx}$dydx | $=$= | $f'\left(x\right)\times n\left[f\left(x\right)\right]^{n-1}$f′(x)×n[f(x)]n−1 |
Multiply the derivative of the inside function by the outside function |
| $=$= | $-3\sin\left(3x\right)\times2\left(\cos\left(3x\right)\right)$−3sin(3x)×2(cos(3x)) |
|
|
| $=$= | $-6\sin\left(3x\right)\cos\left(3x\right)$−6sin(3x)cos(3x) |
Practice questions
Question 4
Differentiate $y=x\tan4x$y=xtan4x.
Question 5
Find the derivative of $\frac{\sin x}{x-8}$sinxx−8.
Question 6
Differentiate $y=\sin^6\left(3x\right)$y=sin6(3x).
When differentiating trigonometric functions it is essential that the argument of the trigonometric function be expressed in radian measure. This requirement comes from the the proof we used to derive the rules which applied limits in radians. To differentiate a function with the argument in degrees we can first rewrite the function, for example we can rewrite the function $y=\sin\left(x^\circ\right)$y=sin(x°) as $y=\sin\left(\frac{\pi x}{180}\right)$y=sin(πx180) and then use the chain rule to differentiate.