What are the first and second derivatives of # g(x) = lnsinx#?

Answer 1

Using chain rule, we can rename #u=sinx# and solve it following chain rule definition: #(dy)/(dx)=(dy)/(du)(du)/(dx)#

#(dy)/(dx)=(1/u)*cosx#
#(dy)/(dx)=cosx/sinx#
This is the same as #tan^(-1)x# or #1/tanx#, which is the same as #cotx#
#(dy^2)/(d^2x)=-csc^2x#
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Answer 2

The first derivative of ( g(x) = \ln(\sin x) ) is ( g'(x) = \frac{\cos x}{\sin x} ).

The second derivative of ( g(x) = \ln(\sin x) ) is ( g''(x) = -\frac{1}{\sin 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.

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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