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How do I find the derivative of #y=arctan(cos θ)#?

Answer 1

Ezra Adams

#y'=-(sintheta)/(cos^(2)theta+1)#

#y=arctan(costheta)#

Applying the chain rule:

#y'=(1)/(cos^(2)theta+1)xx(d[costheta])/("d"theta)#

#=(-sintheta)/(cos^(2)theta+1)#

#=-(sintheta)/(cos^(2)theta+1)#

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Answer 2

Luke Alvarez

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

[ \frac{dy}{d\theta} = \frac{-\sin \theta}{1 + \cos^2 \theta} ]

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Answer 3

Luke Alvarez

To find the derivative of (y = \arctan(\cos \theta)), you can use the chain rule. The chain rule states that if (y = f(g(x))), then (y' = f'(g(x)) \cdot g'(x)). Applying this rule to the given function:

Let (f(u) = \arctan(u)) and (g(\theta) = \cos \theta).
Find the derivatives of (f(u)) and (g(\theta)): [f'(u) = \frac{1}{1 + u^2}] [g'(\theta) = -\sin \theta]
Substitute (g(\theta)) and (g'(\theta)) into the chain rule formula: [y' = f'(g(\theta)) \cdot g'(\theta)]

By combining the derivatives of (f(u)) and (g(\theta)) as per the chain rule, you can find the derivative of (y) with respect to (\theta).

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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.