How do you differentiate $sin^3 4x$ ?

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Answer 1 · 2016-08-02T00:47:56

$d/dx sin^3(4x)=12sin^2(4x)cos(4x)$

Using the chain rule:

$d/dx sin^3(4x) = d/dx(sin(4x))^3$

$=3(sin(4x))^2(d/dxsin(4x))$

$=3sin^2(4x)(cos(4x)(d/dx4x))$

$=3sin^2(4x)cos(4x)(4)$

$=12sin^2(4x)cos(4x)$

How do you differentiate sin^3 4xsin^3 4x?