# How tp prove the derivative of tanx ?

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We know that (tanx)' = sec^2 x

Let us prove this fact.

We know that tanx= sinx/ cosx

Let us assume that f(x) = tanx

Then, f(x) = sinx/ cosx

Then f(x) is a quotient of two functions. Then we will use the quotient rule to determine the first derivative.

Let f(x) = u/ v such that:

u= sinx ==> u' = cosx

v = cosx ==> v' = -sinx

Then we know that:

f'(x) = (u'*v - u*v')/ v^2

= ( cosx*cosx - sinx*-sinx)/ cos^2 x

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

From trigonometric properties, we know that:

sin^2 x + cos^2 x = 1

==> f'(x) = 1/ cos^2 x = ( 1/cosx)^2

But we also know that secx = 1/cosx

**==> f'(x) = sec^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.**