Derivative of tanx=sec^2x, Proof?

Question

Is there a way to prove this using only the definition of the derivative (that is, the difference quotient)?

All help is greatly appreciated.

thank you scott, I eventually got the same answer, although my process was extremely longer.
To those who used the quotient rule - I appreciate your time; however, I must do this by using the DEFINITION OF DERIVATIVE, or DIFFERENCE QUOTIENT.

Thank you all for your time!

Answer 1

Yes.
You need to use the trigonometric identity
tan(a + b) = [tan(a) + tan(b)] / [1 - tan(a)tan(b)]

f '(x) = lim(h→0) [f(x+h)-f(x)]/h

= lim(h→0)[tan(x+h) - tan(x)] / h
= lim(h→0){[tan(x) + tan(h)]/[1 - tan(x)tan(h)] - tan(x)} / h
= lim(h→0)tan(h)[1 + tan^2(x)] / [h(1 - tan(x)tan(h)]
= lim(h→0)tan(h)/h * [1 + tan^2(x)] / [1 - tan(x)tan(h)]

now lim(h→0)tan(h)/h = 1
and lim(h→0)tan(h) = 0
so

f '(x) = lim(h→0) 1*[1 + tan^2(x)] /1 = 1 + tan^2(x) = sec^2(x)

Answer 2

remember that
tan x =sinx/cosx
use the quotient rule, and also the fact
that
sin^2 x + cos^2 x =1
to get
that the derivative of tangent is: 1/cos^2 x =sec^2 x

d tanx /dx
= ( cosx(cosx)-sinx(-sinx) )/cos^2(x)
= 1/cos^2(x)
=sec^2 x

Answer 3

tan x = sin x / cos x
d/dx (tan x) = d/dx (sin x / cosx)
d/dx (sin x / cosx) =
(cosx * cosx - sinx * -sin x) / cos² x =
(cos²x +sin² x) / cos² x
1 / cos²x = sec²x