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

First, recognize that this can be written as

Nasty simplification from here on out, now that the calculus is over...

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

- Identify the outer function as ( \ln(u) ) and the inner function as ( \sqrt{\arcsin(v)} ).
- Compute the derivative of the outer function with respect to its argument ( u ), which is ( \frac{1}{u} ).
- Compute the derivative of the inner function with respect to its argument ( v ), which is ( \frac{1}{2\sqrt{v(1-v)}} ).
- Compute the derivative of the innermost function with respect to ( x ) using the chain rule.
- Substitute the derivatives into the chain rule formula: ( f'(x) = \frac{1}{\sqrt{\arcsin(e^{2-x^2})}} \cdot \frac{1}{2\sqrt{\arcsin(e^{2-x^2})\left(1-\arcsin(e^{2-x^2})\right)}} \cdot \frac{d}{dx}\left(\arcsin(e^{2-x^2})\right) ).

Then, compute ( \frac{d}{dx}\left(\arcsin(e^{2-x^2})\right) ) using the chain rule and simplify the expression.

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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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