Question

How do you find the derivative of `y=tan(x)` ?

Answer 1

`y = tan x`

`dy/dx = sec^2 x`

This is a common identity which many people memorize, along with the derivatives of `sin x` and `cos x` because they appear so frequently.

It can be proven easily using two well-known trig identities and the quotient rule.

Process:
First recall from pre-calculus that `tan theta = sin theta / cos theta`.

So, we can rewrite `y = tan x` equivalently as:

`y = sin x / cos x`

Now, we differentiate, and apply the quotient rule:

`dy/dx = (d/dx[sinx]*cosx - d/dx[cosx]*sinx)/(cos x)^2`

We know that the derivative of `sinx` is `cosx`, and that the derivative of `cosx` is `-sinx`. So, upon simplifying the above equation, we arrive at:

`dy/dx = (cos^2 x + sin^2 x)/(cos^2 x)`

It should be clear that the numerator can be simplified using another trig identity.

Recall `sin^2 theta + cos^2 theta = 1`. (or the Pythagorean identity)

Substitution yields:

`dy/dx = 1/(cos^2 x)`

which is equivalent to:

`dy/dx = sec^2 x`