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#