Question

I want the proof for derivative of Tan( x )

Answer 1

write tan as sin/cos and derive away at it :)

Answer 2

That would be a good way to escape the problem, I am trying to prove it using limits

Answer 3

then it should be the same as the proofs for sin and cos; only with sin/cos

i aint got any idea how to apply thos delta epsilons anymore ...

Answer 4

the limit of a quotient is the quotient of limits ....

Answer 5

Yes the idea is same limits or no limits ... btw no limit is still limits just we are abstracting the whole operation.

Answer 6

with any luck, by limit they mean first principles; but even that is handwavy at times

Answer 7

The difference quotient is
\[ \frac{ \tan x - \tan a}{x - a} \]
The derivative is the limit of this as x --> a. (If this notation bothers you, use the other form of difference quotient where x = a + h and the limit is as h --> 0).

Now, first remember that
\[ \tan x - \tan a = \tan(x-a) . (1 + \tan x . \tan a) \]
thus the difference quotient is equal to
\[ \frac{ \tan(x-a)}{x - a} (1 + \tan x . \tan a) = \frac{ \sin(x-a)}{x-a} \frac{1}{\cos(x-a)} (1 + \tan x . \tan a) \]
Now, the limit of the first term as \( x \rightarrow a \) is 1. The limit of the second term is also 1. The limit of the third term is
\[ 1 + \tan^2 a = \sec^2 a \]
Therefore
\[ \lim_{x \rightarrow a} \frac{ \tan x - \tan a}{x - a} = \sec^2 a \]