Applying Differentiation Rules to Trigonometric Functions

By applying the differentiation rules we have learned so far, we can find the derivatives of trigonometric functions. The differentiation of the six basic trigonometric functions (which are \(\sin, \cos, \tan, \csc, \sec,\) and \(\cot\)) can be done as shown below:

(1) For \(y=\sin x ,\) we use \(\sin a - \sin b = 2\cos \left ( \frac{a+b}{2} \right ) \sin \left ( \frac{a-b}{2} \right)\) to get \( y' :\)

\[ \begin{align} y' &= \lim_{h \rightarrow 0} \frac{\sin(x+h)-\sin x}{h} \\ &= \lim_{h \rightarrow 0} \frac{2\cos \left ( x+ \frac{h}{2} \right) \sin \frac{h}{2}}{h} \\ &= \lim_{h \rightarrow 0} \cos \left(x + \frac{h}{2} \right) \cdot \frac{\sin \frac{h}{2}}{\frac{h}{2}}\\ &= \cos x. & \left(\text{since }\lim_{x\rightarrow0}\frac{\sin x}{x}=1\right) \end{align} \]

(2) For \(y=\cos x ,\) we use \(\cos a - \cos b = -2\sin \left ( \frac{a+b}{2} \right ) \sin \left ( \frac{a-b}{2} \right)\) to get \( y' :\)

\[ \begin{align} y' &= \lim_{h \rightarrow 0} \frac{\cos(x+h)-\cos x}{h} \\ &= \lim_{h \rightarrow 0} \frac{-2\sin \left ( x+ \frac{h}{2} \right) \sin \frac{h}{2}}{h} \\ &= -\lim_{h \rightarrow 0} \sin \left(x + \frac{h}{2} \right) \cdot \frac{\sin \frac{h}{2}}{\frac{h}{2}} \\ &= -\sin x. &\left(\text{since }\lim_{x\rightarrow0}\frac{\sin x}{x}=1\right) \end{align} \]

(3) For \(y= \tan x,\) we convert it to \(\frac{\sin x}{\cos x}\) and use the quotient rule, which gives

\[ \begin{align} y' &= (\tan x)' \\ &= \left ( \frac{\sin x}{\cos x} \right )' \\ &= \frac{\cos x \cdot \cos x- \sin x \cdot (-\sin x)}{\cos^2 x} \\ &= \frac{1}{\cos^2 x} &\left(\text{since }\sin^2 x+\cos^2 x=1\right)\\ &= \sec^2 x. \end{align} \]

(4) For \(y= \cot x,\) we use the same method as for \(y=\tan x,\) which gives

\[ \begin{align} y' &= (\cot x)' \\ &= \left ( \frac{\cos x}{\sin x} \right )' \\ &= \frac{-\sin x \cdot \sin x- \cos x \cdot \cos x}{\sin^2 x} \\ &= \frac{-1}{\sin^2 x} \\ &= -\csc^2 x. \end{align} \]

(5) For \(y= \sec x ,\) we have

\[ \begin{align} y' &= (\sec x)' \\ &= \left (\frac{1}{\cos x} \right )' \\ &= \frac{0-1 \cdot (-\sin x)}{\cos^2 x} \\ &= \frac{\sin x}{\cos^2 x} \\ &= \sec x \cdot \tan x. \end{align} \]

(6) For \(y= \csc x ,\) we have

\[ \begin{align} y' &= (\csc x)' \\ &= \left (\frac{1}{\sin x} \right )' \\ &= \frac{0-1 \cdot (\cos x)}{\sin^2 x} \\ &= \frac{-\cos x}{\sin^2 x} \\ &= -\csc x \cdot \cot x. \end{align} \]

The box below summarizes the derivative of the six trigonometric functions, which of course should be memorized:

1. \(\ y = \sin x \implies y' = \cos x \)
2. \(\ y = \cos x \implies y' = -\sin x \)
3. \(\ y= \tan x \implies y' = \sec^2 x \)
4. \(\ y= \cot x \implies y' = -\csc^2 x \)
5. \(\ y = \sec x \implies y' = \sec x \cdot \tan x \)
6. \(\ y= \csc x \implies y' = -\csc x \cdot \cot x \)

Find the derivative of \(y= \tan x - 3 \cot x .\)

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We have

\[ \begin{align} y' &= \sec^2 x -3\big(-\csc^2 x\big) \\ &= \sec ^2 x + 3\csc^2 x.\ _ \square \end{align} \]

Find the derivative of \( f(x)=\sec^2 x .\)

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By applying the chain rule, we have

\[ \begin{align} f'(x) &= 2\sec x\cdot(\sec x)' \\ &= 2\sec x \cdot \sec x \cdot \tan x \\ &= 2\sec^2 x \cdot \tan x.\ _ \square \end{align} \]

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

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Applying the product rule gives

\[ \begin{align} f'(x) &= (2x^2+1)'\sin 2x + (2x^2+1)(\sin 2x)' \\ &= 4x \sin 2x + (2x^2 +1) \cdot 2\cos 2x \\ &= 4x \sin 2x + 2(2x^2+1) \cos 2x.\ _ \square \end{align} \]

What is \( \frac{dy}{dx} \) in the equation \( \sin x + \sin y = 3?\)

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This equation should be differentiated implicitly. Differentiating both sides with respect to \(x\) gives

\[ \begin{align} \cos x + \cos y \frac{dy}{dx} &= 0 \\ \Rightarrow \frac{dy}{dx} &= -\frac{\cos x}{\cos y}.\ _ \square \end{align} \]

If \( g(x) \) is the inverse function of \( f(x) = \cos x\ \ \left ( 0 < x < \frac{\pi}{2} \right) ,\) then what is \( g'\left( \frac{1}{2} \right) ?\)

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Since \( g(x) \) is the inverse function of \( f(x) ,\) we have

\[ f\big(g(x)\big) = x \implies \cos g(x) = x.\]

Differentiating both sides of \( \cos g(x) = x\) gives

\[ \begin{align} \big( - \sin g(x) \big) \cdot g'(x) &= 1 \\ \Rightarrow g'(x) &= -\frac{1}{\sin g(x)}. \qquad (1) \end{align} \]

From \( \cos \frac{\pi}{3}=\frac{1}{2},\) we know that \( g \left( \frac{1}{2} \right) = \frac{\pi}{3}.\) Plugging this into (1) gives

\[ \begin{align} g'\left( \frac{1}{2} \right) &= -\frac{1}{\sin g \left( \frac{1}{2} \right)} \\ &= -\frac{1}{\sin \frac{\pi}{3}} \\ &= -\frac{1}{\hspace{3mm} \frac{\sqrt{3}}{2}\hspace{3mm} } \\ &= - \frac{2\sqrt{3}}{3}.\ _ \square \end{align} \]

If \( f(x) = \sin^2 \frac{x}{2} ,\) what is \( {\displaystyle\lim_{x \rightarrow \pi}} \frac{f'(x)}{x-\pi}?\)

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We have

\[ \begin{align} f'(x) &= 2\sin \frac{x}{2} \cdot \cos \frac{x}{2} \cdot \frac{1}{2} \\ &= \frac{1}{2} \sin x &\qquad (\text{since } \sin 2A = 2 \sin A \cdot \cos A )\\ \Rightarrow f'(\pi) &= 0. \end{align} \]

It follows that

\[ \begin{align} \lim_{x \rightarrow \pi} \frac{f'(x)}{x-\pi} &= \lim_{x \rightarrow \pi} \frac{f'(x) - f'(\pi)}{x-\pi} \\ &= f''(\pi). &\qquad (1) \end{align} \]

Since \( f''(x) = \frac{1}{2} \cos x ,\) from \( (1)\) we have

\[ \begin{align} \lim_{x \rightarrow \pi}\frac{f'(x)}{x-\pi} &= f''(\pi) \\ &= \frac{1}{2} \cos \pi \\ &= - \frac{1}{2}.\ _ \square \end{align} \]