# tan x derivativeHow to prove the derivative of tan x ?

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You also may use the following formula for tangent function, such that:

`tan x = sqrt(1 - cos^2 x)/(cos x)`

You need to find derivative of `tan x` using the quotient rule and chain rule, such that:

`(tan x)' = ((sqrt(1 - cos^2x))'*cos x - sqrt(1 - cos^2 x)*(cos x)')/(cos^2 x)`

`(tan x)' = ((-2cos x*(-sin x))/(2sqrt(1 - cos^2x))*cos x - sqrt(1 - cos^2 x)*(- sin x))/(cos^2 x)`

`(tan x)' = ((2cos x*(sin x))/(2sin x)*cos x + sin x*(sin x))/(cos^2 x)`

Reducing duplicate factors yields:

`(tan x)' = (cos^2 x + sin^2 x)/(cos^2 x)`

Using the basic trigonometric formula yields:

`cos^2 x + sin^2 x = 1`

`(tan x)' = 1/(cos^2 x) = 1 + tan^2 x`

**Hence, evaluating the derivative of tangent function, yields **`(tan x)' = 1 + tan^2 x.`

We'll start from the formula of the derivative of the tangent function:

(tan x)' = 1/(cos x)^2

We know the fact that the tangent function is a ratio of the functions sine and cosine.

(tan x) = sin x/cos x

If we'll differentiate with respect to x, we'll get:

(tan x)' = {[d/dx(sin x)]*cos x - sin x*[d/dx(cos x)]}/(cos x)^2

d/dx(sin x) = cos x

d/dx(cos x) = -sin x

We'll substitute d/dx(sin x) and d/dx(cos x) by their results and we'll get:

(tan x)' = [cos x*cos x - sin x*(-sin x)]/(cos x)^2

(tan x)' = [(cos x)^2 + (sin x)^2]/(cos x)^2

But, from the fundamental formula of trigonometry, we'll get:

(cos x)^2 + (sin x)^2 = 1

(tan x)' = 1/(cos x)^2 q.e.d.