[Calculus Series Part 3] Week 2: Trig Derivatives and the Chain Rule

This post is for students who already worked through the Week 1 limit proofs and now want to turn those results into usable derivative rules. Week 2 of calculus focuses on trigonometric differentiation and the chain rule for composite functions. This is where the limits and addition formulas from the previous week become concrete derivative formulas such as $(\sin x)'$, $(\cos x)'$, and $(\tan x)'$. The aim is to follow the flow from the definition of the derivative to those formulas, not simply recite them.

1. Differentiating trigonometric functions

1.1 $(\sin x)' = \cos x$

Trig differentiation starts from the limit definition. Last week we already proved $\lim_{h \to 0}\frac{\sin h}{h}=1$ and $\lim_{h \to 0}\frac{1-\cos h}{h}=0$. Those two limits are exactly what lets the derivative calculation finish cleanly.

$$\frac{d}{dx} \sin x = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}.$$

Now substitute the angle-addition formula

$$\sin(x+h)=\sin x\cos h + \cos x\sin h$$

into the difference quotient:

$$\frac{\sin(x+h)-\sin x}{h} = \frac{(\sin x\cos h + \cos x\sin h)-\sin x}{h}.$$

Group the $\sin x$ terms and split the fraction:

$$\frac{\sin(x+h)-\sin x}{h} = \frac{\sin x(\cos h-1)+\cos x\sin h}{h} = \sin x \cdot \frac{\cos h -1}{h} + \cos x \cdot \frac{\sin h}{h}.$$

Taking limits kills the first term and leaves the second term, so

$$\frac{d}{dx} \sin x = \cos x.$$

Geometrically, this says the y-coordinate on the unit circle changes at a rate of $\cos x$ when the angle changes slightly. That picture matches the limit calculation, but the derivative formula is justified by the algebra above.

1.2 $(\cos x)' = -\sin x$

Cosine follows the same script, now using $\lim_{h \to 0}\frac{\cos h -1}{h}=0$ explicitly.

$$\frac{d}{dx} \cos x = \lim_{h \to 0} \frac{\cos(x+h)-\cos x}{h}.$$

Use the angle-addition formula

$$\cos(x+h)=\cos x \cos h - \sin x \sin h,$$

and substitute it into the difference quotient:

$$\frac{\cos(x+h)-\cos x}{h} = \frac{(\cos x \cos h - \sin x \sin h)-\cos x}{h}.$$

After regrouping,

$$\frac{\cos(x+h)-\cos x}{h} = \cos x \cdot \frac{\cos h -1}{h} - \sin x \cdot \frac{\sin h}{h}.$$

Taking limits yields

$$\frac{d}{dx} \cos x = -\sin x.$$

Geometrically, sine and cosine are rotated versions of each other, so one derivative naturally becomes the negative of the other.

1.3 Core trig derivative table

These formulas become reference points for everything that follows. Functions like $\sec x$ or $\csc x$ are best re-derived from sine and cosine via the quotient rule or reciprocals.

$f(x)$	$f'(x)$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$
$\cot x$	$-\csc^2 x$
$\sec x$	$\sec x \tan x$
$\csc x$	$-\csc x \cot x$

These formulas hold on the parts of the domain where each original function is defined.

Group them mentally:

• $\sin x$ and $\cos x$ form a pair.
• Differentiating cosine flips the sign.
• Tangent and secant stay linked via $\sec^2 x$ and $\sec x \tan x$.
• Cotangent and cosecant usually introduce a minus sign.

If you want to see where that tangent-secant link comes from, re-derive $(\sec x)'$ from $\sec x=\frac{1}{\cos x}$:

$$(\sec x)' = \left(\frac{1}{\cos x}\right)' = \frac{(0)(\cos x) - (1)(-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{\sin x}{\cos x}\cdot\frac{1}{\cos x} = \tan x\sec x.$$

That is why secant and tangent keep appearing together in the derivative table.

2. Quotient rule and chain rule

2.1 Quotient rule

For functions $u(x)$ and $v(x)$,

$$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}.$$

Applying it to

$$\tan x = \frac{\sin x}{\cos x}$$

gives

$$(\tan x)' = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x.$$

The key identity in the middle is $\sin^2 x + \cos^2 x = 1$. The most common mistakes are forgetting to square the denominator or reversing $u'v-uv'$. If needed, verbalize it: "differentiate the top times the bottom minus the top times the derivative of the bottom, over the bottom squared."

Another example:

$$f(x)=\frac{\cos x}{x}$$

leads to

$$f'(x)=\frac{(-\sin x)\cdot x - (\cos x)\cdot 1}{x^2} = \frac{-x\sin x - \cos x}{x^2}.$$

Computing a few of these by hand makes later mixed problems far less intimidating.

2.2 Chain rule

For a composition $f(g(x))$,

$$\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x).$$

Think of it as differentiating the outer function first, then multiplying by the derivative of the inner function. Examples:

• $\sin(3x)$ → outer $\sin u$, inner $u=3x$.
• $(2x^2+1)^4$ → outer $u^4$, inner $u=2x^2+1$.
• $e^{\sin x}$ → outer $e^u$, inner $u=\sin x$.

Sample computations:

$$\frac{d}{dx}\sin(2x) = 2\cos(2x),$$ $$\frac{d}{dx}(x^2+1)^5 = 10x(x^2+1)^4,$$ $$\frac{d}{dx}e^{\sin x} = e^{\sin x}\cos x.$$

All three share the same backbone: differentiate the outer layer, then multiply by the inner derivative.

Before moving on, keep two common chain-rule mistakes in view:

• $\frac{d}{dx}\sin(3x)=\cos(3x)$ is incomplete because it forgets the inner derivative $3$.
• $\frac{d}{dx}(x^2+1)^5=5(x^2+1)^4$ is also incomplete because it forgets the inner derivative $2x$.

In both cases, the outer derivative is only the first half of the work.

2.3 Choosing between quotient rule and chain rule

Students often ask which rule should come first. A quick test helps:

• If one function is wrapped around another, as in $\sin(x^2+1)$, start with the chain rule.
• If two separate functions are divided, as in $\frac{\sin x}{x^2+1}$, start with the quotient rule.
• If both happen at once, differentiate the inside pieces correctly and then assemble the larger rule.

2.4 Trig problems plus the chain rule

We rarely differentiate trig functions with a bare $x$ inside. Typical expressions look like

• $\sin(5x-1)$
• $\cos(x^2)$
• $\tan(\sqrt{x})$.

For example,

$$\frac{d}{dx}\cos(x^2) = -\sin(x^2)\cdot 2x,$$

A frequent misunderstanding is to stop at $-\sin(x^2)$ and forget the $2x$. That means the outer derivative was found, but the inner derivative was skipped. Memorizing trig derivatives alone only gets you halfway; multiplying by the inner derivative completes the job.

3. Quick checks

Checkpoint 1

Differentiate $f(x) = \sin\left(3x + \frac{\pi}{4}\right)$.

Show answer

$$f'(x) = 3\cos\left(3x + \frac{\pi}{4}\right).$$

The factor $3$ comes from differentiating the inside expression $3x+\frac{\pi}{4}$.

Checkpoint 2

Find $f'(x)$ for $f(x) = (2x^2 + 1)^4$.

Show answer

Chain rule gives

$$f'(x) = 4(2x^2 + 1)^3 \cdot 4x = 16x(2x^2 + 1)^3.$$

Stopping at $4(2x^2+1)^3$ would mean forgetting the derivative of the inside.

Checkpoint 3

Differentiate $f(x) = \frac{\sin x}{x^2 + 1}$.

Show answer

Let $u = \sin x$ and $v = x^2 + 1$. Then $u' = \cos x$ and $v' = 2x$, so

$$f'(x)=\frac{u'v-uv'}{v^2} = \frac{\cos x(x^2+1)-\sin x(2x)}{(x^2+1)^2}.$$

Checkpoint 4

Differentiate $f(x)=e^{\sin x}$.

Show answer

Outer function $e^u$, inner $u=\sin x$, so

$$f'(x)=e^{\sin x}\cos x.$$

Checkpoint 5

Differentiate $f(x)=\cos(x^2+1)$.

Show answer

Applying the chain rule,

$$f'(x)=-\sin(x^2+1)\cdot 2x = -2x\sin(x^2+1).$$

4. Frequent mistakes

• Dropping the minus sign in $(\cos x)'=-\sin x$.
• Writing $(\tan x)'=\sec x$ instead of $\sec^2 x$.
• Reversing $u'v-uv'$ in the quotient rule.
• Stopping after differentiating the outer layer in chain-rule problems.
• Differentiating $\sin(3x)$ but rewriting it as $3\cos x$ and losing the entire inside.

5. Key takeaway

This week tied trig derivatives to the chain rule. The main lesson is not the table itself, but recognizing where each ingredient appears: addition formulas and limits justify the base derivatives, the quotient rule extends them, and the chain rule makes them usable in real problems.

In short:

• Week 1 limits built the Week 2 trig derivatives.
• Sine and cosine derivatives set up tangent (and friends).
• Once the chain rule joins in, we can tackle genuine exam-style expressions.

So treat this week as practice in understanding how formulas arise and in which order to apply them, rather than rote memorization.