# How do you differentiate #y=sin(arccost)#?

We will use the chain rule.

so

so

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To differentiate ( y = \sin(\arccos t) ), use the chain rule:

[ \frac{dy}{dt} = \frac{d}{dt}(\sin(\arccos t)) = \frac{d}{dt}(\sin(\theta)) ]

where ( \theta = \arccos t ). Applying the chain rule:

[ \frac{dy}{dt} = \cos(\arccos t) \cdot \frac{d}{dt}(\arccos t) = \cos(\arccos t) \cdot \frac{-1}{\sqrt{1 - t^2}} ]

[ \frac{dy}{dt} = \cos(\arccos t) \cdot \frac{-1}{\sqrt{1 - t^2}} = -\frac{1}{\sqrt{1 - t^2}} ]

Therefore, the derivative of ( y = \sin(\arccos t) ) is ( -\frac{1}{\sqrt{1 - t^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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