So far, we have built a solid foundation for finding derivatives of polynomials and exponential functions. Now, we turn our attention to another essential class of functions that model periodic behavior everywhere in science and engineering: the trigonometric functions. How do we find the rate of change of a sine wave at a particular instant?
Before we can differentiate trig functions, let's ensure we remember what they are and their values at key angles. The six trigonometric functions are defined based on a right triangle with angle $\theta$, an opposite side (opp), an adjacent side (adj), and a hypotenuse (hyp).
It is crucial to have these common values memorized. They will appear frequently in examples and on exams.
| $\theta$ | $0$ | $\pi/6$ | $\pi/4$ | $\pi/3$ | $\pi/2$ | $\pi$ | $3\pi/2$ |
|---|---|---|---|---|---|---|---|
| $\sin(\theta)$ | $0$ | $1/2$ | $\sqrt{2}/2$ | $\sqrt{3}/2$ | $1$ | $0$ | $-1$ |
| $\cos(\theta)$ | $1$ | $\sqrt{3}/2$ | $\sqrt{2}/2$ | $1/2$ | $0$ | $-1$ | $0$ |
For a more in-depth review, please see the "Precalculus Review" link under Modules/Resources on our course Canvas page.
Let's build some intuition. What should the derivative of $f(x) = \sin(x)$ look like? Remember, the derivative represents the slope of the tangent line. Let's examine the graph of $\sin(x)$ and estimate its slope at several points.
If we plot these slope values—$1, 0, -1, 0$—at $x=0, \pi/2, \pi, 3\pi/2$, we get points that fall exactly on the graph of $g(x) = \cos(x)$. This suggests a remarkable relationship: the derivative of the sine function appears to be the cosine function.
Our visual inspection strongly suggests that $\frac{d}{dx}(\sin x) = \cos x$. To prove this formally, we must return to the limit definition of the derivative. This proof relies on two fundamental trigonometric limits.
The following two limits are foundational for the derivatives of all trigonometric functions. We will take them as given, but their proofs can be found in the textbook.
You can convince yourself of these limits by exploring the following interactive graph.
Let $f(x) = \sin(x)$. We use the limit definition $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
$$ \frac{d}{dx}(\sin x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h} $$We use the sum identity for sine: $\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)$.
$$ = \lim_{h \to 0} \frac{(\sin x \cos h + \cos x \sin h) - \sin(x)}{h} $$Now, we rearrange the terms by grouping the $\sin(x)$ terms together.
$$ = \lim_{h \to 0} \frac{(\sin x \cos h - \sin x) + \cos x \sin h}{h} $$ $$ = \lim_{h \to 0} \left( \frac{\sin x (\cos h - 1)}{h} + \frac{\cos x \sin h}{h} \right) $$We can factor out the terms that don't depend on $h$ from the limits.
$$ = \sin(x) \left( \lim_{h \to 0} \frac{\cos h - 1}{h} \right) + \cos(x) \left( \lim_{h \to 0} \frac{\sin h}{h} \right) $$Using our two special limits, the first limit is $0$ and the second is $1$.
$$ = \sin(x) \cdot (0) + \cos(x) \cdot (1) $$ $$ = \cos(x) $$Thus, we have formally proven that $\frac{d}{dx}(\sin x) = \cos x$. A similar derivation shows that $\frac{d}{dx}(\cos x) = -\sin x$.
Now that we know the derivatives of sine and cosine, we can find the derivative of tangent using the Quotient Rule.
$$ \frac{d}{dx}(\tan x) = \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) $$Let $f(x) = \sin x$ (high) and $g(x) = \cos x$ (low). Then $f'(x) = \cos x$ and $g'(x) = -\sin x$.
$$ = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2} = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{(\cos x)^2} $$ $$ = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} $$Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:
$$ = \frac{1}{\cos^2 x} = \sec^2 x $$| Function | Derivative |
|---|---|
| $f(x) = \sin(x)$ | $f'(x) = \cos(x)$ |
| $f(x) = \cos(x)$ | $f'(x) = -\sin(x)$ |
| $f(x) = \tan(x)$ | $f'(x) = \sec^2(x)$ |
The derivatives of the remaining three trigonometric functions (cosecant, secant, and cotangent) can all be found using the Quotient Rule as well. The full list is provided below.
Here are the derivatives of all six trigonometric functions. These must be memorized.
| $\frac{d}{dx}(\sin x) = \cos x$ | $\frac{d}{dx}(\cos x) = -\sin x$ |
| $\frac{d}{dx}(\tan x) = \sec^2 x$ | $\frac{d}{dx}(\cot x) = -\csc^2 x$ |
| $\frac{d}{dx}(\sec x) = \sec x \tan x$ | $\frac{d}{dx}(\csc x) = -\csc x \cot x$ |
Notice a pattern: The derivative of every "co-" function (cosine, cotangent, cosecant) is negative.
Differentiate $f(x) = x^2 \sin(x)$.
We use the Product Rule with $g(x)=x^2$ and $h(x)=\sin(x)$. Then $g'(x)=2x$ and $h'(x)=\cos(x)$.
$$ f'(x) = g'(x)h(x) + g(x)h'(x) $$ $$ f'(x) = (2x)(\sin x) + (x^2)(\cos x) = 2x\sin x + x^2\cos x $$Find the derivative of $y = \frac{\cos x}{1 - \sin x}$.
The position of a mass on a spring is given by the equation $s(t) = 5\cos(t)$, where $s$ is in centimeters and $t$ is in seconds. Find the velocity and acceleration of the mass at time $t$. What is the acceleration at $t=\pi/3$?
The velocity is the derivative of the position function.
$$ v(t) = s'(t) = \frac{d}{dt}(5\cos t) = 5(-\sin t) = -5\sin t $$The acceleration is the derivative of the velocity function.
$$ a(t) = v'(t) = \frac{d}{dt}(-5\sin t) = -5(\cos t) = -5\cos t $$Now, we find the acceleration at $t=\pi/3$ seconds.
$$ a(\pi/3) = -5\cos(\pi/3) = -5 \left(\frac{1}{2}\right) = -2.5 \text{ cm/s}^2 $$After working through the calculations, the following interactive graph can help visualize the motion of the mass, its velocity, and its acceleration over time.
Find the derivative of $g(t) = t^3\cos(t) - \sin(t)$.
Find the equation of the tangent line to the curve $y = \sec(x)$ at the point $(\pi/4, \sqrt{2})$.
Find the derivative of $y = e^x \tan(x)$.
Today we have dramatically expanded our differentiation toolkit by adding rules for all six trigonometric functions. We saw how these rules can be derived from the limit definition and how they combine with our previous rules (Product and Quotient) to tackle more complex functions. These derivatives are fundamental in physics and engineering for modeling any kind of wave or oscillation.
After this lecture, you should be able to: