3.5.1 Derivatives of Trigonometric Functions

Fundamental Trigonometric Limits

• Two "gateway" limits are used to build all derivative formulas for trigonometric functions.
  • $\displaystyle \lim_{x \to 0} \frac{\sin x}{x}=1$
  • $\displaystyle \lim_{x \to 0} \frac{\cos x-1}{x}=0$
• Direct substitution fails (denominator $\to 0$), but each result can be justified by:
  • Numerical tables
  • Graphs of $\sin x$ and $\cos x$ zoomed near the origin
  • Geometry of the unit circle (arclength vs. chord)

Using the Gateway Limits to Evaluate Other Limits

• Strategy: Rewrite a target limit so that the pattern $\sin( ext{something})/(\text{same something})$ or $(\cos(\text{something})-1)/(\text{same something})$ appears.
• Example 1: $\displaystyle \lim_{x \to 0} \frac{\sin(4x)}{4}$
  • Multiply numerator & denominator by $4$ so that the argument and denominator match.
  • $\frac{\sin(4x)}{4}=\frac{4\sin(4x)}{4\cdot4x}=\frac{\sin(4x)}{4x}$
  • Substitute $t=4x$ (so $t\to0$)
  • $4\cdot \lim_{t\to0}\frac{\sin t}{t}=4\cdot1=4$
• Example 2: $\displaystyle \lim_{x \to 0} \frac{\sin(3x)}{\sin(5x)}$
  • Divide numerator & denominator by $x$:
  • $\frac{\sin(3x)}{\sin(5x)}=\frac{ \big(\sin(3x)/(3x)\big)\,3}{ \big(\sin(5x)/(5x)\big)\,5}$
  • Each bracketed ratio $\to1$, so the whole limit $=3/5$.

Derivatives of the Basic Trigonometric Functions

• Using the definition of the derivative together with the two gateway limits, one proves (details in textbook):
  • $\displaystyle \frac{d}{dx}[\sin x]=\cos x$
  • $\displaystyle \frac{d}{dx}[\cos x]=-\sin x$
• Geometric intuition:
  • $\sin x$ and $\cos x$ are periodic ➔ their slopes (derivatives) should also oscillate.

Differentiation Examples

Useful rules: Product rule $\big(uv\big)'=u'v+uv'$; Quotient rule $\big(\tfrac{u}{v}\big)'=\dfrac{u'v-uv'}{v^{2}}$ ; Sum/Difference rule $[u\pm v]'=u'\pm v'$.

(a) $f(x)=e^{x}\cos x$

• Product rule
  • $f'=\big(e^{x}\big)'\cos x+e^{x}(\cos x)'$
  • $=e^{x}\cos x+e^{x}(-\sin x)$
  • Factor common term: $f'(x)=e^{x}\big(\cos x-\sin x\big)$

(b) $g(x)=\sin x - x\cos x$

• Treat as a difference; second term needs product rule.
  1. Derivative of $\sin x$ is $\cos x$.
  2. For $x\cos x$: $1\cdot\cos x + x(-\sin x)$.
• Assemble:
  • $g'=\cos x-\big(\cos x - x\sin x\big)=x\sin x$

(c) $h(x)=\dfrac{1+\sin x}{1-\sin x}$

• Quotient rule with $u=1+\sin x,\;u'=\cos x$ and $v=1-\sin x,\;v'=-\cos x$.
  • $h'=\dfrac{u'v-uv'}{v^{2}}=\dfrac{\cos x(1-\sin x)-(1+\sin x)(-\cos x)}{(1-\sin x)^{2}}$
  • Expand the numerator:
  • First term: $\cos x - \cos x\sin x$
  • Second term: $+(1+\sin x)\cos x$
  • Cancellation: $-\cos x\sin x + \cos x\sin x =0$
  • Remaining: $2\cos x$
  • Result: $h'(x)=\dfrac{2\cos x}{\big(1-\sin x\big)^{2}}$

Connections & Take-Aways

• Mastery of the two fundamental limits is essential; they recur in almost every proof involving trigonometric derivatives.
• Because all other trig functions ($\tan,\cot,\sec,\csc$) can be written with $\sin$ and $\cos$, their derivatives ultimately rely on these basic formulas.
• Algebraic tactics (multiplying/dividing by matching factors, clever substitutions) are often the fastest path to limit evaluation and simplification.
• Periodicity carries over: derivatives of periodic functions generally remain periodic.
• Practical implication: Trigonometric derivatives appear in physics (harmonic motion), engineering (signal processing), and any subject dealing with waves.