# How do you differentiate #f(x) = ln(sqrt(arcsin(e^(2-x)) ) # using the chain rule?

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Just adding to Alexander's answer,

Also, for differentiation, arc sine should be taken as a single-valued function..

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To differentiate the function ( f(x) = \ln\left(\sqrt{\arcsin(e^{2-x})}\right) ) using the chain rule, follow these steps:

- Recognize the composition of functions within the expression: ( \ln(\sqrt{u}) ) where ( u = \arcsin(e^{2-x}) ).
- Differentiate the outer function with respect to the inner function: ( \frac{d}{du}(\ln(u)) = \frac{1}{u} ).
- Differentiate the inner function with respect to ( x ).
- Use the chain rule: ( \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} ).

Now, let's break down the differentiation:

- ( \frac{d}{du}(\ln(u)) = \frac{1}{u} )
- ( \frac{d}{dx}(\arcsin(e^{2-x})) = \frac{1}{\sqrt{1 - (e^{2-x})^2}} \cdot \frac{d}{dx}(e^{2-x}) )
- ( \frac{d}{dx}(e^{2-x}) = -e^{2-x} )
- Putting it all together:

[ \frac{df}{dx} = \frac{1}{\sqrt{\arcsin(e^{2-x})}} \cdot \frac{1}{\sqrt{1 - (e^{2-x})^2}} \cdot (-e^{2-x}) ]

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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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