# How do you differentiate #y=sqrt[(x^2-1)/(x^2+1)]#?

So using the power rule and chain rule twice we find:

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To differentiate the function ( y = \sqrt{\frac{x^2 - 1}{x^2 + 1}} ), we'll use the chain rule and the quotient rule.

Let ( u = \frac{x^2 - 1}{x^2 + 1} ), then ( y = \sqrt{u} ).

Using the chain rule, the derivative of ( y ) with respect to ( x ) is:

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

Now, let's find ( \frac{du}{dx} ) using the quotient rule:

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

[ = \frac{2x(x^2 + 1 - x^2 + 1)}{(x^2 + 1)^2} ]

[ = \frac{2x(2)}{(x^2 + 1)^2} ]

[ = \frac{4x}{(x^2 + 1)^2} ]

Substitute ( \frac{du}{dx} = \frac{4x}{(x^2 + 1)^2} ) back into the expression for ( \frac{dy}{dx} ):

[ \frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{4x}{(x^2 + 1)^2} ]

[ = \frac{2x}{(x^2 + 1)^2 \sqrt{\frac{x^2 - 1}{x^2 + 1}}} ]

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

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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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