What is the derivative of #f(x)=cos(-x)-cos(x)#?

1 Answer
Nov 3, 2015

It's just a tricky way to write the identitcal-zero function, so the derivative is zero.

Explanation:

I'd say that no derivatives are needed, since you know that #cos(-x)=cos(x)#, and so

#cos(-x)=cos(x) = cos(x)-cos(x) = 0#, and the derivative of #0# is of course #0#.

Anyway, even if we didn't notice that, we can do the derivatives: since the derivative of #cos(x)# is #-sin(x)#, we have (using the chain rule for the first term)

#-sin(-x) * (d/dx (-x)) - (-sin(x))#

and since #(d/dx (-x))=-1#, the expression becomes

#sin(-x) +sin(x)#

Again, you should use the fact that #sin(-x)=-sin(x)#, and so the sum is again zero.