Question

If `f(x) = sin x`, what is `f''(pi/6)`?

Answer 1

`-1/2`

Explanation:

The derivative of `sin(x)` and its sequential derivatives form a pattern:

`{(d/dx(sinx)=cosx),(d/dx(cosx)=-sinx),(d/dx(-sinx)=-cosx),(d/dx(-cosx)=sinx):}`

And the process repeats, forming a cycle of `4`. (Very similar to the powers of `i`...)

Anyway, the second derivative of `sinx` is `-sinx`, since

`f(x)=sinx`
`f'(x)=cosx`
`f''(x)=-sinx`

Thus,

`f''(x)=-sin(pi/6)=-1/2`

Answer 2

Here a another kind of answer

`d^n/dx^nsin(x) = sin(x + npi/2)`

here `n = 2` and `x = pi/6`

`d^2/dx^2sin(pi/6) = sin(pi/6+pi) = sin((7pi)/6) = -1/2`