How to Find The Derivative of a Trigonometric Function

Derivative of \( sin(x) \) :

\( \begin{align*}
\frac{d}{dx} \sin x &= \lim_{h \to 0} \frac{\sin(x + h) – \sin x}{h} \\
&= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h – \sin x}{h} \\
&= \lim_{h \to 0} \left(\sin x \cdot \frac{\cos h – 1}{h} + \cos x \cdot \frac{\sin h}{h}\right) \\
&= \sin x \cdot \lim_{h \to 0} \frac{\cos h – 1}{h} + \cos x \cdot \lim_{h \to 0} \frac{\sin h}{h} \\
&= \sin x \cdot 0 + \cos x \cdot 1 \\
&= \cos x
\end{align*}\)

So the derivative of \(sin (x)\), is \(cos (x)\). Now, using chain rule, we can have:

\( sin(f(x))=cos (f(x))×f’ (x) \)

Derivative of \(cos (x) \):

Almost the same procedure can be used to find the derivative of \( cos (x) \) :

\(\begin{align*}
\frac{d}{dx} \cos x &= \lim_{h \to 0} \frac{\cos(x + h) – \cos x}{h} \\
&= \lim_{h \to 0} \frac{\cos x \cos h – \sin x \sin h – \cos x}{h} \\
&= \lim_{h \to 0} \left(\cos x \cdot \frac{\cos h – 1}{h} – \sin x \cdot \frac{\sin h}{h}\right) \\
&= \cos x \cdot \lim_{h \to 0} \frac{\cos h – 1}{h} – \sin x \cdot \lim_{h \to 0} \frac{\sin h}{h} \\
&= \cos x \cdot 0 – \sin x \cdot 1 \\
&= -\sin x
\end{align*}\)

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Derivative of \(tan (x) \) and \(cot (x) \):

Now we could do the same thing for tangent and cotangent, but it would be easier to use quotient rule and find them using  \( \tan x = \frac{\sin x}{\cos x} \) and  \( \cot x = \frac{\cos x}{\sin x} \):

\( \frac{d}{dx} \tan x = \frac{d}{dx} \left( \frac{\sin x}{\cos x} \right) = \frac{\cos x \cdot \cos x – \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x \)

And:

\( \frac{d}{dx} \cot x = \frac{d}{dx} \left( \frac{\cos x}{\sin x} \right) = \frac{\sin x \cdot (-\sin x) – \cos x \cdot \cos x}{\sin^2 x} = \frac{-\sin^2 x – \cos^2 x}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x \)