The two Special Trigonometric Limits were the hard part. With them, the derivatives of sine and cosine take a few lines, and the The Quotient Rule gives the other four trig functions.
🔑 Key idea
$$\frac{d}{dx}\sin x = \cos x, \qquad \frac{d}{dx}\cos x = -\sin x,$$ $$\frac{d}{dx}\tan x = \sec^2 x, \quad \frac{d}{dx}\sec x = \sec x\tan x, \quad \frac{d}{dx}\csc x = -\csc x\cot x, \quad \frac{d}{dx}\cot x = -\csc^2 x.$$ ($x$ in radians.) The "co" functions all get a minus sign.
💡 Why it works
Use $\sin(x + h) = \sin x\cos h + \cos x\sin h$: $$\frac{\sin(x + h) - \sin x}{h} = \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h}.$$ As $h \to 0$, the special limits give $\sin x \cdot 0 + \cos x \cdot 1 = \cos x$. The same steps with $\cos(x + h) = \cos x\cos h - \sin x\sin h$ give $-\sin x$. For tangent, the quotient rule: $$\left(\frac{\sin x}{\cos x}\right)' = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$$.
The slopes of $y = \sin x$ form the graph of $y = \cos x$: steepest upward at $x = 0$ (slope 1), flat at $x = \frac{\pi}{2}$ (slope 0).On the unit circle, $\cos\theta$ is greatest exactly when $\sin\theta$ is rising fastest, and $\cos$ is falling where $\sin$ is near its top.
✏️ Worked example: A sum
Differentiate $3\sin x - 2\cos x$.
Differentiate each term: $\frac{d}{dx}(3\sin x) = 3\cos x$ and $\frac{d}{dx}(-2\cos x) = -2(-\sin x)$.
The two minus signs make a plus.
Answer: $3\cos x + 2\sin x$
✏️ Worked example: A product
Differentiate $x^2\sin x$.
Product rule: $(x^2)'\sin x + x^2(\sin x)'$.
$= 2x\sin x + x^2\cos x$.
Answer: $2x\sin x + x^2\cos x$
✏️ Worked example: A slope at a special angle
Find the slope of $y = \sin x + \cos x$ at $x = \frac{\pi}{3}$.
$y' = \cos x - \sin x$.
At $\frac{\pi}{3}$: $\frac{1}{2} - \frac{\sqrt{3}}{2}$.
Answer: $\frac{1 - \sqrt{3}}{2} \approx -0.366$
Why radians? The limit $\frac{\sin h}{h} \to 1$ needs radians. If $x$ were measured in degrees, every derivative would pick up a factor $\frac{\pi}{180}$: the slope of $\sin(x^\circ)$ is $\frac{\pi}{180}\cos(x^\circ)$. Calculus uses radians so that this factor is exactly 1.
A memory aid for the other four: tangent and secant derivatives are built from tangent and secant ($\sec^2 x$, $\sec x\tan x$); their "co" partners are built from cotangent and cosecant, with a minus sign.
⚠️ Common mistake
Dropping the minus sign: writing $\frac{d}{dx}\cos x = \sin x$. Look at the graph of $\cos x$ just to the right of 0: it is falling, so its derivative must be negative there, but $\sin x$ is positive. Check any trig derivative at one easy point this way.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.