# How do you differentiate # g(x) = sqrtarccos(x^2-1) #?

Use implicit differentiation and the chain rule to find that

With that, the remainder of the problem can be done using the chain rule:

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To differentiate ( g(x) = \sqrt{\arccos(x^2-1)} ), we can use the chain rule:

Let ( u = \arccos(x^2-1) ). Then ( y = \sqrt{u} ).

Using the chain rule: [ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} ]

First, find ( \frac{du}{dx} ): [ \frac{du}{dx} = \frac{d}{dx}[\arccos(x^2-1)] ]

Applying the chain rule again: [ \frac{du}{dx} = \frac{-1}{\sqrt{1 - (x^2-1)^2}} \cdot \frac{d}{dx}(x^2-1) ]

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

Now, find ( \frac{dy}{du} ): [ \frac{dy}{du} = \frac{d}{du}[\sqrt{u}] ]

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

Combine both: [ \frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{-1}{\sqrt{1 - (x^2-1)^2}} \cdot 2x ]

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

Finally, substitute ( u = \arccos(x^2-1) ): [ \frac{dy}{dx} = \frac{-x}{\sqrt{\arccos(x^2-1)(1 - (x^2-1)^2)}} ]

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