What is the derivative of #y=ln(cos^2ɵ)#?

Answer 1

Assuming that we want #y' = dy/(d theta)#, in simplest form: #y' = -2tan theta#

#y = ln(cos^2 theta)#

Method 1 Leave it as is and use the chain rule twice:

#y' = 1/cos^2 theta * d/(d theta) (cos^2 theta)#
#= 1/cos^2 theta * 2cos theta * d/(d theta) (cos theta)#
#= 1/cos^2 theta * 2cos theta * d/(d theta) (cos theta)#
#= 1/cos^2 theta * 2cos theta * (-sin theta)#
# = -2 sintheta/costheta = -2tan theta#
Method 2 Use properties of #ln# to rewrite:
#y = ln(cos^2 theta) = 2ln(cos theta)#

Use the chain rule: (less detail this time)

#y' = 2*1/cos theta *(-sin theta) = -2tan theta#
For derivative with respect to #x#
#dy/dx = dy/(d theta)* (d theta)/dx#

So

#dy/dx = -2tan theta (d theta)/dx#
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Answer 2

To find the derivative of ( y = \ln(\cos^2\theta) ), you can use the chain rule. The derivative is:

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

Simplified, this becomes:

[ \frac{dy}{d\theta} = \frac{2\sin\theta}{\cos\theta} ]

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Answer 3

The derivative of ( y = \ln(\cos^2 \theta) ) with respect to ( \theta ) is ( \frac{d}{d\theta}[\ln(\cos^2 \theta)] = \frac{-2\sin \theta}{\cos^2 \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.

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.

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.

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.

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