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

Answer 1

Cameron Alexander

#=-(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))#

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Answer 2

Aurora Alvarado

To find the derivative of ( f(x) = \arctan(\cos(3x)) ), you can use the chain rule and the derivative of the arctan function. The chain rule states that if ( u ) is a function of ( x ), and ( v ) is a function of ( u ), then the derivative of their composition is given by ( \frac{dv}{dx} = \frac{dv}{du} \cdot \frac{du}{dx} ). Applying this rule:

[ \frac{d}{dx}[\arctan(\cos(3x))] = \frac{1}{1+(\cos(3x))^2} \cdot (-\sin(3x)) \cdot \frac{d}{dx}(3x) ]

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

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.