# Differentiation of cosx

Here you will learn what is the differentiation of cosx and its proof by using first principle.

Let’s begin –

## Differentiation of cosx

The differentiation of cosx with respect to x is -sinx.

i.e. $$d\over dx$$ (cosx) = -sinx

## Proof Using First Principle :

Let f(x) = cos x. Then, f(x + h) = cos(x + h)

$$\therefore$$   $$d\over dx$$(f(x)) = $$lim_{h\to 0}$$ $$f(x + h) – f(x)\over h$$

$$d\over dx$$(f(x)) = $$lim_{h\to 0}$$ $$cos(x + h) – cos x\over h$$

By using trigonometry formula,

[cos C – cos D = $$-2sin{C + D\over 2}sin{C – D\over 2}$$]

$$d\over dx$$(f(x)) = $$lim_{h\to 0}$$ $$-2sin({h\over 2})sin({{2x + h}\over 2})\over h$$

$$d\over dx$$(f(x)) = $$lim_{h\to 0}$$ $$-2sin({h/2})sin({{x + h/2}\over 2})\over 2(h/2)$$

$$\implies$$ $$d\over dx$$(f(x)) = -$$lim_{h\to 0}$$ $$sin({{x + h/2}\over 2})$$ $$lim_{h\to 0}$$$$sin(h/2)\over (h/2)$$

because, [$$lim_{h\to 0}$$$$sin(h/2)\over (h/2)$$ = 1]

$$\implies$$ $$d\over dx$$(f(x)) = -(sin x) $$\times$$ 1 = – sin x

Hence, $$d\over dx$$ (cos x) = -sin x

Example : What is the differentiation of cos x – 2 sin x with respect to x?

Solution : Let y = cos x – 2 sin x

$$d\over dx$$(y) = $$d\over dx$$(cos x – 2 sin x)

$$\implies$$ $$d\over dx$$(y) = $$d\over dx$$(cos x) – $$d\over dx$$(2 sinx)

By using cosx and sinx differentiation we get,

$$\implies$$ $$d\over dx$$(y) = -sin x – 2 $$d\over dx$$(sinx)

$$\implies$$ $$d\over dx$$(y) = -sin x – 2 cos x

Hence, $$d\over dx$$(cos x – 2 sin x) = -sin x – 2 cos x

Example : What is the differentiation of $$x^2$$ +  cos 2x with respect to x?

Solution : Let y = $$x^2$$ + cos 2x

$$d\over dx$$(y) = $$d\over dx$$($$x^2$$ +  cos 2x)

$$\implies$$ $$d\over dx$$(y) = $$d\over dx$$$$x^2$$ – $$d\over dx$$(cos 2x)

By using chain rule and differentiation formulas we get,

$$\implies$$ $$d\over dx$$(y) = 2x + (2)(-sin 2x)

Hence, $$d\over dx$$($$x^2$$ +  cos 2x) = 2x – 2sin 2x

### Related Questions

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