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

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Answer 1 · 2017-05-29T10:48:54

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

$d/dx(tan^-1(cos(3x)))$

Step 1. Use the chain rule
$d/dx(tan^-1(cos(3x)))=(d(tan^-1(u)))/(du)(du)/(dx)$ ,

where $u=cos(3x)$ and $d/(du)(tan^-1(u))=1/(1+u^2)$

$=(d/dx(cos(3x)))/(1+cos^2(3x))$

Step 2. Using the chain rule again,

$d/dx(cos(3x))=(d(cos(u)))/(du)(du)/(dx)$ ,

where $u=3x$ and $d/(du)(cos(u))-sin(u)$ , gives

$=(-d/dx(3x)sin(3x) )/(1+cos^2(3x))$

Step 3. Factor out constants

$=-3(sin(3x) )/(1+cos^2(3x))d/dx(x)$

ANSWER: $=-(3sin(3x) )/(1+cos^2(3x))$

How do you find the derivative of f(x) = arctan(cos(3x))f(x) = arctan(cos(3x))?