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

Question

Aurora Alvarado · Accepted Answer

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

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.

Paisley Johnson · Answer

undefined The derivative of $f(x) = \cos(-x) - \cos(x)$ is $f'(x) = \sin(-x) + \sin(x)$.

Aurora Alvarado · Answer

undefined The derivative of $ f(x) = \cos(-x) - \cos(x) $ is $ f'(x) = \sin(x) + \sin(-x) $, which simplifies to $ f'(x) = \sin(x) - \sin(x) $. Therefore, the derivative of $ f(x) = \cos(-x) - \cos(x) $ is $ f'(x) = 0 $.