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

2 Answers
Apr 1, 2015

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]}

Apr 1, 2015

#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#