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,

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

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