How do you find the derivative of Inverse trig function $y = sin cos(3x -2)$ ?

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Answer 1 · 2015-09-25T01:21:11

To find the derivative of $y = sin (cos(3x -2))$ , use the chain rule twice:

$dy/dx = cos(cos(3x -2))d/dx(cos(3x -2))$

$= cos(cos(3x -2))[-sin(3x-2)d/dx(3x-2)]$

$= cos(cos(3x -2))[-sin(3x-2)(3)]$

$= -3sin(3x-2)cos(cos(3x -2))$

How do you find the derivative of Inverse trig function y = sin cos(3x -2)y = sin cos(3x -2)?