# How do you find the differential #dy# of the function #y=xsqrt(1-x^2)#?

The differential of

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To find the differential (dy) of the function (y = x \sqrt{1 - x^2}), we can use the product rule and the chain rule of differentiation. The product rule states that if (y = u \cdot v), where (u) and (v) are functions of (x), then (dy = u \cdot dv + v \cdot du).

In this case, let (u = x) and (v = \sqrt{1 - x^2}). Then, (du = dx) and (dv = \frac{-x}{\sqrt{1 - x^2}} , dx).

Now, applying the product rule:

[ \begin{aligned} dy &= u \cdot dv + v \cdot du \ &= x \cdot \frac{-x}{\sqrt{1 - x^2}} , dx + \sqrt{1 - x^2} \cdot dx \ &= -\frac{x^2}{\sqrt{1 - x^2}} , dx + \sqrt{1 - x^2} , dx \ &= \left(\sqrt{1 - x^2} - \frac{x^2}{\sqrt{1 - x^2}}\right) , dx \ &= \frac{1 - x^2 - x^2}{\sqrt{1 - x^2}} , dx \ &= \frac{1 - 2x^2}{\sqrt{1 - x^2}} , dx. \end{aligned} ]

So, the differential (dy) of the function (y = x \sqrt{1 - x^2}) is (dy = \frac{1 - 2x^2}{\sqrt{1 - x^2}} , dx).

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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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