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How do you find the derivative of #y = arctan(cosx)#?

Answer 1

Cameron Alexander

In this way, with the chain rule:

#y'=1/(1+cos^2x)*(-sinx)#.

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Answer 2

Evelyn Agard

To find the derivative of ( y = \arctan(\cos(x)) ), use the chain rule:

( \frac{dy}{dx} = \frac{d}{dx}(\arctan(\cos(x))) = \frac{1}{1+(\cos(x))^2} \cdot \frac{d}{dx}(\cos(x)) )

Now, differentiate ( \cos(x) ) with respect to ( x ), which is ( -\sin(x) ):

( \frac{dy}{dx} = \frac{-\sin(x)}{1+(\cos(x))^2} )

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Answer 3

Ezekiel Alexander

To find the derivative of ( y = \arctan(\cos(x)) ), you can use the chain rule. The derivative is:

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

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

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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.