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

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To differentiate ( y = \sin^{-1}(-2x^2) ), we'll use the chain rule and the derivative formula for ( \sin^{-1}x ), which is ( \frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}} ).

Given ( y = \sin^{-1}(-2x^2) ),

Let ( u = -2x^2 ), so ( y = \sin^{-1}(u) ).

First, differentiate ( u = -2x^2 ) with respect to ( x ):

[ \frac{du}{dx} = -4x ]

Then, using the derivative formula for ( \sin^{-1}u ), we have:

[ \frac{dy}{du} = \frac{1}{\sqrt{1-u^2}} ]

Substituting ( u = -2x^2 ) into this formula gives:

[ \frac{dy}{du} = \frac{1}{\sqrt{1-(-2x^2)^2}} = \frac{1}{\sqrt{1-4x^4}} ]

Finally, using the chain rule ( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} ), we get:

[ \frac{dy}{dx} = \frac{1}{\sqrt{1-4x^4}} \cdot (-4x) = \frac{-4x}{\sqrt{1-4x^4}} ]

Therefore, the derivative of ( y = \sin^{-1}(-2x^2) ) with respect to ( x ) is ( \frac{-4x}{\sqrt{1-4x^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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