What is the derivative of #sin(x(pi/8))#?

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Answer 1 · 2018-01-10T03:42:27

#y'=d/dx[sin((pix)/8)]=pi/8cos((pix)/8)#

Apply the Chain Rule:

#y'=d/dx[f(g(x)]=f'(g(x))*g'(x)#

Given: #sin(x(pi/8))#, We can rewrite this as #sin((pix)/8)#

Let #f(x)=sinx# and #g(x)=(pix)/8#

Thus, #f'(x)=cosx# and #g'(x)=pi/8#

So,

#y'=d/dx[sin((pix)/8)]=cos((pix)/8)*pi/8#