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How do you use the chain rule to differentiate #ln(-cosx)#?

Answer 1

Christopher Allers

#d/dxln(-cosx)=-tanx#

The Chain Rule, when applied to logarithms, tells us that

#d/dxln(u)=1/lnu*(du)/dx, u# is some function in terms of #x.# Here, #u=-cosx,# so we have

#d/dxln(-cosx)=1/(-cosx)*d/dx(-cosx)#

#d/dxln(-cosx)=sinx/(-cosx)#

#d/dxln(-cosx)=-tanx#

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Answer 2

Ezekiel Alexander

To differentiate ln(-cosx) using the chain rule, follow these steps:

Recognize that ln(-cosx) can be rewritten as ln(u), where u = -cosx.
Apply the chain rule, which states that the derivative of ln(u) with respect to x is (1/u) * du/dx.
Compute du/dx by differentiating u = -cosx with respect to x. The derivative of -cosx is sinx.
Substitute u = -cosx and du/dx = sinx into the chain rule formula.
The derivative of ln(-cosx) with respect to x is (1/(-cosx)) * sinx.

Therefore, the derivative of ln(-cosx) with respect to x is (sinx) / (cosx).

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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.