Differentiation of Trigonometric Functions

Before any formula, the derivative of $\sin x$ can be read directly from the curve. The slope of $y = \sin x$ is exactly what $y = \cos x$ describes at each point.

Reading the slope of $y = \sin x$:

• At $x = 0$: the sine curve crosses zero and is rising steeply — slope is at its maximum, +1. And $\cos(0) = 1$. Match!
• At $x = \tfrac{\pi}{2}$: the curve is at its peak — it flattens to a slope of 0. And $\cos(\tfrac{\pi}{2}) = 0$. Match!
• At $x = \pi$: the curve crosses zero and is falling steeply — slope is at its minimum, -1. And $\cos(\pi) = -1$. Match!
• At $x = \tfrac{3\pi}{2}$: the curve is at its trough — it flattens again, slope = 0. And $\cos(\tfrac{3\pi}{2}) = 0$. Match!

Reading the slope of $y = \cos x$: starts flat (slope 0 at $x=0$), then decreases (negative slope) reaching minimum at $x = \tfrac{\pi}{2}$. This matches $-\sin x$ exactly.

$\dfrac{d}{dx}\sin x = \cos x$ — read from the graph: slope of sin matches the cos curve; $\dfrac{d}{dx}\cos x = -\sin x$ — the negative sign: cos starts flat then slopes down

Pause — copy the two foundational trig derivatives $\dfrac{d}{dx}\sin x = \cos x$ and $\dfrac{d}{dx}\cos x = -\sin x$ — radians only — into your book.

We just saw that $\dfrac{d}{dx}\sin x = \cos x$ and $\dfrac{d}{dx}\cos x = -\sin x$. That raises a question: $\tan x = \dfrac{\sin x}{\cos x}$ — so can we derive its derivative from these two results? This card answers it → applying the quotient rule to $\dfrac{\sin x}{\cos x}$, then using $\sin^2 x + \cos^2 x = 1$ to arrive at $\sec^2 x$.

Since $\tan x = \dfrac{\sin x}{\cos x}$, we can derive its derivative using the quotient rule — the same rule from earlier in this topic.

Full derivation:

Key steps: apply quotient rule with $u = \sin x$, $v = \cos x$; then use the Pythagorean identity $\cos^2 x + \sin^2 x = 1$.

• Result: $\dfrac{d}{dx}\tan x = \sec^2 x$ where $\sec x = \dfrac{1}{\cos x}$
• Domain: defined everywhere except $x = \dfrac{\pi}{2} + k\pi$ (where $\cos x = 0$)
• Sign: $\sec^2 x = \dfrac{1}{\cos^2 x} \geq 1 > 0$ always — so $\tan x$ is always increasing on its domain
• HSC tip: write $\sec^2 x$, not $\dfrac{1}{\cos^2 x}$ or $1 + \tan^2 x$ (all equivalent, but $\sec^2 x$ is the standard form)

$\dfrac{d}{dx}\tan x = \sec^2 x$ — derived via quotient rule on $\tfrac{\sin x}{\cos x}$; Proof steps: quotient rule → $\dfrac{\cos^2 x + \sin^2 x}{\cos^2 x}$ → use identity $\sin^2+\cos^2=1$ → $\dfrac{1}{\cos^2 x} = \sec^2 x$

Pause — copy the result $\dfrac{d}{dx}\tan x = \sec^2 x$ and its proof sketch (quotient rule on $\frac{\sin x}{\cos x}$, then Pythagorean identity) into your book.

We just saw the three basic trig derivatives: $\sin x \to \cos x$, $\cos x \to -\sin x$, $\tan x \to \sec^2 x$. That raises a question: what happens when the argument is $ax + b$ instead of just $x$? This card answers it → the chain rule multiplies each result by $a$, the derivative of the inner linear function.

The chain rule — differentiate the outer function, keep the inner, multiply by the inner derivative — extends all three trig rules to any composite expression. For linear inner functions $ax + b$, the pattern is clean and systematic.

Chain rule for trig: Let $u = ax + b$, so $\dfrac{du}{dx} = a$.

Pattern: derivative of outer trig function (with inner unchanged) × derivative of inner function. The $a$ multiplies out the front.

Three worked examples:

• $\dfrac{d}{dx}\sin(3x) = 3\cos(3x)$  —  outer: sin → cos; inner derivative: 3
• $\dfrac{d}{dx}\cos(2x - \pi) = -2\sin(2x - \pi)$  —  outer: cos → -sin; inner derivative: 2
• $\dfrac{d}{dx}\tan(5x) = 5\sec^2(5x)$  —  outer: tan → sec²; inner derivative: 5

$\dfrac{d}{dx}\sin(ax+b) = a\cos(ax+b)$ — the $a$ comes from the inner derivative; $\dfrac{d}{dx}\cos(ax+b) = -a\sin(ax+b)$ — the negative is from the outer (cos → -sin), the $a$ from the inner

Pause — copy the three chain rule forms $\dfrac{d}{dx}\sin(ax+b) = a\cos(ax+b)$, $\dfrac{d}{dx}\cos(ax+b) = -a\sin(ax+b)$, $\dfrac{d}{dx}\tan(ax+b) = a\sec^2(ax+b)$ into your book.

We just saw that for linear inner functions $ax+b$, the chain rule simply multiplies by $a$. That raises a question: what if the inner function is $x^2$, $\sqrt{x}$, or $e^x$ — where the inner derivative is no longer a constant? This card answers it → the general chain rule: $\dfrac{d}{dx}\sin(f(x)) = f'(x)\cos(f(x))$, and the same pattern for cos and tan.

When the inner function is more complex than $ax + b$ — a polynomial, an exponential, a square root — the chain rule still applies. The outer trig rule stays the same; the inner derivative just gets more interesting.

General forms:

Three examples with non-linear inner functions:

• $\dfrac{d}{dx}\sin(x^2)$: inner $u = x^2$, $u' = 2x$. Answer: $2x\cos(x^2)$
• $\dfrac{d}{dx}\cos(\sqrt{x})$: inner $u = x^{1/2}$, $u' = \dfrac{1}{2\sqrt{x}}$. Answer: $-\dfrac{\sin(\sqrt{x})}{2\sqrt{x}}$
• $\dfrac{d}{dx}\tan(e^x)$: inner $u = e^x$, $u' = e^x$. Answer: $e^x\sec^2(e^x)$

Step-by-step method:

1. Identify the outer trig function (sin, cos, or tan)
2. Write down the inner function $u = f(x)$ and find $u' = f'(x)$
3. Write the derivative of the outer trig function (with inner unchanged)
4. Multiply by $f'(x)$

$\dfrac{d}{dx}\sin(f(x)) = f'(x)\cos(f(x))$ — the $f'(x)$ multiplier is the chain rule; Example: $\dfrac{d}{dx}\sin(x^2) = 2x\cos(x^2)$ — inner derivative $2x$ multiplies out front

Pause — copy the general chain rule $\dfrac{d}{dx}\sin(f(x)) = f'(x)\cos(f(x))$ and the worked example $\dfrac{d}{dx}\sin(x^2) = 2x\cos(x^2)$ into your book.

We just saw how to differentiate any composite trig expression using the general chain rule. That raises a question: how do you deploy these derivatives in HSC problem types — finding tangent lines, computing rates in oscillating systems, and solving $y' = k$? This card answers it → a systematic method using point-slope form $y - y_1 = m(x - x_1)$ after substituting the differentiated trig expression.

Trig derivatives unlock three families of HSC problems: finding tangent (and normal) lines at points on trig curves, computing rates of change in oscillating systems, and locating where gradients equal a given value.

1. Tangent to a trig curve. Find the equation of the tangent to $y = \sin(2x)$ at $x = \dfrac{\pi}{4}$.

2. Rate of change — oscillating system. A pendulum has displacement $x(t) = 0.1\cos(2t)$ metres. Find its velocity and speed at $t = \dfrac{\pi}{3}$.

3. Where does the gradient equal a given value? Solve $\dfrac{d}{dx}[3\sin x] = \dfrac{3\sqrt{2}}{2}$ for $x \in [0, 2\pi]$.

Tangent to trig curve: differentiate, substitute $x$-value to get slope $m$, use $y - y_1 = m(x - x_1)$; Horizontal tangent: solve $y' = 0$ — for $\sin/\cos$ this means solving a trig equation

Pause — copy the tangent method (differentiate, substitute $x$-value for slope $m$, apply $y - y_1 = m(x-x_1)$) and the horizontal tangent condition (solve $y' = 0$) into your book.