How do you find the derivatives of inverse function $f(x) = arctan(cos(3x))$ ?

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Answer 1 · 2015-06-17T01:25:11

$d/(dx)[arctan(u(v(x)))] = 1/(1+u^2)*((du)/(dv))((dv)/(dx))$

$d/(dx)[cosu] = -sinu*((du)/(dx))$

With $u(v) = cosv$ and $v(x) = 3x$ :

$d/(dx)[arctan(cos3x)] = 1/(1+cos^2(3x))*-sin(3x)*3$

$= (-3sin3x)/(1+cos^2(3x))$

How do you find the derivatives of inverse function f(x) = arctan(cos(3x))f(x) = arctan(cos(3x))?