# How do you differentiate #y=sin^-1(2x+1)#?

Therefore, by Chain Rule,

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To differentiate ( y = \sin^{-1}(2x + 1) ), use the chain rule:

- Differentiate the outer function (\sin^{-1}(u)) with respect to its inner function (u), which is (2x + 1). This derivative is ( \frac{1}{\sqrt{1 - u^2}} ).
- Then, differentiate the inner function (2x + 1) with respect to (x), which yields (2).

So, the derivative of ( y = \sin^{-1}(2x + 1) ) is ( \frac{2}{\sqrt{1 - (2x + 1)^2}} ).

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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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