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

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

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Answer 1 · 2017-05-29T10:48:54

$= - \frac{3 sin (3 x)}{1 + {cos}^{2} (3 x)}$ $=-(3sin(3x) )/(1+cos^2(3x))$

Explanation:

$\frac{d}{d x} ({tan}^{- 1} (cos (3 x)))$ $d/dx(tan^-1(cos(3x)))$

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

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

$= \frac{\frac{d}{d x} (cos (3 x))}{1 + {cos}^{2} (3 x)}$ $=(d/dx(cos(3x)))/(1+cos^2(3x))$

Step 2. Using the chain rule again,

$\frac{d}{d x} (cos (3 x)) = \frac{d (cos (u))}{d u} \frac{d u}{d x}$ $d/dx(cos(3x))=(d(cos(u)))/(du)(du)/(dx)$ ,

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

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

Step 3. Factor out constants

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

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

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

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

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Explanation:

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