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Where the derivative of sine comes from, the rest of the table, the cycle of four, and why radians are not optional.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to differentiate all six trigonometric functions, derive $\sin' = \cos$ from the angle-addition formula and the limit proved by the squeeze theorem, combine trigonometric derivatives with the product, quotient and chain rules, keep the minus sign that the co-functions carry, use the cycle of four to find a high-order derivative without computing every step, and say what would change if angles were measured in degrees.
$$\frac{d}{dx}\sin x = \cos x, \qquad \frac{d}{dx}\cos x = -\sin x.$$
Everything else follows. Writing $\tan x = \dfrac{\sin x}{\cos x}$ and using the quotient rule gives $\sec^2 x$; the same move gives the other three:
| $f$ | $f'$ |
|---|---|
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^2 x$ |
| $\cot x$ | $-\csc^2 x$ |
| $\sec x$ | $\sec x \tan x$ |
| $\csc x$ | $-\csc x \cot x$ |
The pattern is worth noticing rather than memorising six lines: the three "co-" functions carry the minus sign, and each of their derivatives is the mirror of its partner's.
Where $\sin' = \cos$ comes from. Apply the definition and the angle-addition formula: $$\frac{\sin(x+h) - \sin x}{h} = \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h}.$$ The two limits from lesson 9 finish it: $\dfrac{\cos h - 1}{h} \to 0$ and $\dfrac{\sin h}{h} \to 1$, leaving $\cos x$.
So this table rests entirely on the squeeze theorem, and entirely on radians — the limit $\sin h / h \to 1$ is a fact about radians, and in degrees it is $\pi/180$.
Almost every use needs the chain rule. $\sin(3x)$, $\cos(x^2)$, $\tan(1/x)$ are all composites, and the inner derivative is a factor.
Another way: picture
Sketch $\sin x$ and read its slopes: steepest upward at $0$ (slope $1$), flat at $\pi/2$ (slope $0$), steepest downward at $\pi$ (slope $-1$). Plot those slopes against $x$ and the cosine curve appears. The derivative of sine is not a formula to remember; it is the slope graph of a curve you already know.
Another way: steps
Differentiating sine four times returns it: $$\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x.$$
So the $n$-th derivative depends only on $n \bmod 4$, and a question about the $100$th derivative is a question about the remainder $0$.
The cycle is not a curiosity. Two steps give $\dfrac{d^2}{dx^2}\sin x = -\sin x$, so sine and cosine solve $$y'' + y = 0,$$ the equation of every undamped oscillation — a mass on a spring, a pendulum through small angles, an LC circuit, a vibrating string. That one differential equation is why these two functions appear everywhere in physics and not merely in triangles, and the cycle of four is where it comes from.
Working in degrees. A calculator in degree mode gives wrong derivatives, and the answers are wrong by the factor $\pi/180 \approx 0.0175$, which is small enough to look like a rounding problem rather than a mistake.
Losing the minus on cosine. $(\cos x)' = -\sin x$. The three co-functions all carry a minus, and it is the single most frequently dropped sign in the subject.
Forgetting the inner derivative. $(\sin 3x)' = 3\cos 3x$. Dropping the $3$ gives an answer that does not depend on the frequency, which is visibly wrong once you think about what the graph does.
Reading $\sin^2 x$ as $\sin(x^2)$. The first is $(\sin x)^2$ with derivative $2\sin x\cos x$; the second is a composite with derivative $2x\cos(x^2)$. The notation is genuinely ambiguous and the convention has to be known.
Quoting $(\tan x)' = \sec^2 x$ where $\cos x = 0$. The derivative exists only where the function does, and $\tan$ has vertical asymptotes.
These six derivatives are usually met as a list to memorise, and memorising them works until something needs to be reconstructed. Two facts generate the whole table: $\lim_{h\to 0}\frac{\sin h}{h} = 1$ and the angle-addition formula. From those, $\sin' = \cos$; from $\sin' = \cos$ and the identity $\cos x = \sin(\pi/2 - x)$, the chain rule gives $\cos' = -\sin$; from those two and the quotient rule, everything else.
This matters when a formula is misremembered — is it $\sec x\tan x$ or $\sec^2 x\tan x$? — and it matters more when the same question is asked about $\sinh$ or about an inverse trigonometric function, where there is no list to have memorised and the derivation is all there is.
$f(x) = \dfrac{\sin x}{x}$. Quotient rule: $u = \sin x$, $v = x$, so $u' = \cos x$ and $v' = 1$.
The outermost operation is division.
$f'(x) = \dfrac{x\cos x - \sin x}{x^2}$.
Numerator's derivative first.
Or rewrite as $x^{-1}\sin x$ and use the product rule: $-x^{-2}\sin x + x^{-1}\cos x$, which is the same expression over a common denominator. Two routes, one answer — and a free check.
$f(x) = x^2\cos(3x)$. Outermost is a product: $u = x^2$, $v = \cos(3x)$.
Frame first.
$u' = 2x$. For $v'$ the chain rule applies: outer $\cos$, inner $3x$, so $v' = -\sin(3x) \cdot 3 = -3\sin(3x)$.
A rule inside a rule.
$f'(x) = 2x\cos(3x) - 3x^2\sin(3x)$. Both the minus sign and the $3$ came from $v'$, and both are commonly lost.
A composite: outer $\sin u$, inner $u = x^2$.
Not a product, not a power of a sine.
Outer derivative at the inside: $\cos(x^2)$. Inner derivative: $2x$.
$f'(x) = 2x\cos(x^2)$. Compare $\sin^2 x$, whose derivative is $2\sin x\cos x$ — a different function entirely.
Match each trigonometric form to its derivative structure.
| Derivative $\cos x$ | Derivative $-\sin x$ | Derivative $2\sec^2(2x)$ | Derivative $\sec x\tan x$ | |
|---|---|---|---|---|
| $\sin x$ | ||||
| $\cos x$ | ||||
| $\tan(2x)$ | ||||
| $\sec x$ |
Match each trigonometric expression to the derivative feature that must appear.
| Derivative is $\cos x$ | Derivative begins with $-\sin x$ | Derivative has a factor $4\sec^2(4x)$ | Derivative is $-\csc x\cot x$ | |
|---|---|---|---|---|
| $\sin x$ | ||||
| $\cos x$ | ||||
| $\tan(4x)$ | ||||
| $\csc x$ |
$f(x) = \sin x$, so $f'(x) = \cos x$. What is $f'(0.7)$, to four decimal places? (Radians.)
Answer:
$f(x) = \sin(7x)$. What is $f'(0)$?
Answer:
Differentiate $f(x) = 2x\sin x$.
$f(x) = \sin x$. What is the $32$-th derivative of $f$ at $x = 0$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
If angles were measured in degrees, what would the derivative of $\sin x$ be?
You can differentiate trigonometric expressions with the other rules in play, and find a high-order derivative from the cycle. Say in your own words why $\sin' = \cos$ is a fact about radians. Next: implicit differentiation, and the derivative of an inverse.
8. Your turn: differentiate $f(x) = \sin(x^2)$, step 3