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

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Answer 1 · 2016-01-09T18:08:21

$-1/2$

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 · 2016-01-09T19:00:53

Here a another kind of answer

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

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

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

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