first principals and derivatives

Question

Let f(x) = sin x. Which of the following would you use to calculate f'(−π / 2) using the definition of derivative (i.e., first principles)?

Answer 1

If you used the definition of a derivative to calculate f'(-π/2), it would look like this:

f'(x) = lim_h->0 [f(x+h)-f(x)]/h

f'(-π/2) = lim_h->0 [sin(-π/2+h)-sin(-π/2)]/h

f'(-π/2) = lim_h->0 [sin(-π/2)cos(h)+cos(-π/2)sin(h)-(-1)]/h

f'(-π/2) = lim_h->0 [-cos(h)+1]/h

f'(-π/2) = lim_h->0 sin(h) <-- L'Hopital's Rule

f'(-π/2) = lim_h->0 sin(0)

f'(-π/2) = 0

Therefore, the derivative of sin(x) at x=-π/2 is 0. This is also simply calculated by knowing f'(x)=cos(x), so cos(-π/2) = 0.

Hope this helped!