How tp prove the derivative of tanx ?

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