Question

How do you evaluate the derivative of the function `f (x) = tan x cot x` at the indicated point (1,1)?

Answer 1

I would use the Product Rule to evaluate the derivative knowing that:
derivative of `tan(x)` is `1/cos^2(x)`
Derivative of `cot(x)` is `-1/sin^2(x)`

(you can find these remembering that `tan(x)=sin(x)/cos(x)` and `cot(x)=cos(x)/sin(x)` and deriving using the Quotient Rule)

So you get:
`f'(x)=1/cos^2(x)cot(x)+tan(x)[-1/sin^2(x)]=0`

Evaluated at `x=1` gives you:
`f'(1)=0`

As an exercise on derivation is good, but you can see from the start that your function can be rearranged as: `f(x)=sin(x)/cos(x)*cos(x)/sin(x)=1` ! Which is a constant with derivative always equal to zero.
graph{tan(x)cot(x) [-10, 10, -5, 5]}

Answer 2

`f(x)=tanx cot x = sinx/cosx cosx/sinx = 1`

`f'(x)=0` for all `x` in the domain of `f`.

Therefore: `f'(1) = 0`