# How do you differentiate #arcsin(csc(1-1/x^3)) )# using the chain rule?

As a Real valued function this is not differentiable.

That is:

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To differentiate ( \arcsin(\csc(1-\frac{1}{x^3})) ) using the chain rule, we first note that the derivative of ( \arcsin(u) ) with respect to ( u ) is ( \frac{1}{\sqrt{1-u^2}} ).

Next, we differentiate ( \csc(v) ) with respect to ( v ) to get ( -\csc(v) \cot(v) ).

Now, let ( u = \csc(1-\frac{1}{x^3}) ) and ( v = 1-\frac{1}{x^3} ). Then, by the chain rule, the derivative of ( \arcsin(\csc(1-\frac{1}{x^3})) ) with respect to ( x ) is:

[ \frac{d}{dx} \arcsin(\csc(1-\frac{1}{x^3})) = \frac{d}{dx} \arcsin(u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} ]

[ = \frac{1}{\sqrt{1-\csc^2(1-\frac{1}{x^3})}} \cdot \frac{d}{dx} \csc(1-\frac{1}{x^3}) ]

[ = \frac{1}{\sqrt{1-\csc^2(v)}} \cdot \frac{d}{dx} \csc(v) ]

[ = \frac{1}{\sqrt{1-\csc^2(1-\frac{1}{x^3})}} \cdot (-\csc(1-\frac{1}{x^3}) \cot(1-\frac{1}{x^3})) \cdot \frac{d}{dx} (1-\frac{1}{x^3}) ]

[ = \frac{1}{\sqrt{1-\csc^2(1-\frac{1}{x^3})}} \cdot (-\csc(1-\frac{1}{x^3}) \cot(1-\frac{1}{x^3})) \cdot \frac{3}{x^4} ]

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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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