# Differentiation of tanx

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

Let’s begin –

## Differentiation of tanx

The differentiation of tanx with respect to x is $$sec^2x$$.

i.e. $$d\over dx$$ (tanx) = $$sec^2x$$

## Proof Using First Principle :

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

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

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

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

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

By using trigonometry formula,

[sin A cos B – cos A sin B = sin (A – B)]

$$\implies$$ $$d\over dx$$(f(x)) = $$lim_{h\to 0}$$ $$sin h\over h$$.$$1\over cos x cos (x + h)$$

$$\implies$$ $$d\over dx$$(f(x)) = $$lim_{h\to 0}$$ $$sin h\over h$$ $$lim_{h\to 0}$$$$1\over cos x cos (x + h)$$

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

$$\implies$$ $$d\over dx$$(f(x)) = 1.$$1\over cos x cos x$$ = $$sec^2x$$

Hence, $$d\over dx$$ (tan x) = $$sec^2x$$

Example : What is the differentiation of tan x – x with respect to x?

Solution : Let y = tan x – x

$$d\over dx$$(y) = $$d\over dx$$(tan x – x)

$$\implies$$ $$d\over dx$$(y) = $$d\over dx$$(tan x) – $$d\over dx$$(x)

By using tanx differentiation we get,

$$\implies$$ $$d\over dx$$(y) = $$sec^2x$$ – 1

Hence, $$d\over dx$$(tan x – x) = $$sec^2x$$ – 1

Example : What is the differentiation of $$tan\sqrt{x}$$ with respect to x?

Solution : Let y = $$tan\sqrt{x}$$

$$d\over dx$$(y) = $$d\over dx$$($$tan\sqrt{x}$$)

By using chain rule we get,

$$\implies$$ $$d\over dx$$(y) = $$1\over 2\sqrt{x}$$$$sec^2\sqrt{x}$$

Hence, $$d\over dx$$($$tan\sqrt{x}$$) = $$1\over 2\sqrt{x}$$$$sec^2\sqrt{x}$$

### Related Questions

What is the Differentiation of tan inverse x ?

What is the Differentiation of sec x ?

What is the Integration of Tan x ?