# How do you compute the derivative of #y=(sin^-1)x# and x=3/5?

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To compute the derivative of ( y = (\sin^{-1} x) ) at ( x = \frac{3}{5} ), you can use the chain rule. The derivative of ( \sin^{-1} x ) is ( \frac{1}{\sqrt{1 - x^2}} ). Applying the chain rule, the derivative of ( y ) with respect to ( x ) is:

[ \frac{d}{dx} (\sin^{-1} x) = \frac{d}{dx} (\arcsin x) = \frac{1}{\sqrt{1 - x^2}} ]

So, at ( x = \frac{3}{5} ), the derivative is:

[ \frac{1}{\sqrt{1 - (\frac{3}{5})^2}} = \frac{1}{\sqrt{1 - \frac{9}{25}}} = \frac{1}{\sqrt{\frac{25}{25} - \frac{9}{25}}} = \frac{1}{\sqrt{\frac{16}{25}}} = \frac{1}{\frac{4}{5}} = \frac{5}{4} ]

Therefore, the derivative of ( y = (\sin^{-1} x) ) at ( x = \frac{3}{5} ) is ( \frac{5}{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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