# How to Find The Derivative of a Trigonometric Function

Using the limit formula to derive functions, we can uncover the general derivative form of trigonometric functions. This process requires an understanding of trigonometric identities. By applying these identities within the limit framework, it becomes possible to systematically determine the derivatives of various trigonometric functions, enhancing calculus applications.

## 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*}

## 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:

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

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