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.