# How do you differentiate #x^2/(sqrt(x+2))-sqrt(x+2)/x^2#?

Please see below.

Use the quotient rule to find

So

Simplify further to taste.

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To differentiate the expression ( \frac{x^2}{\sqrt{x+2}} - \frac{\sqrt{x+2}}{x^2} ), apply the quotient rule:

- Differentiate the numerator and denominator separately.
- Then apply the quotient rule formula: ( \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} ), where ( f(x) ) is the numerator and ( g(x) ) is the denominator.

Differentiating ( \frac{x^2}{\sqrt{x+2}} ):

( f(x) = x^2 ) ( g(x) = \sqrt{x+2} ) ( f'(x) = 2x ) ( g'(x) = \frac{1}{2\sqrt{x+2}} )

Differentiating ( \frac{\sqrt{x+2}}{x^2} ):

( f(x) = \sqrt{x+2} ) ( g(x) = x^2 ) ( f'(x) = \frac{1}{2\sqrt{x+2}} ) ( g'(x) = 2x )

Now, substitute these values into the quotient rule formula:

( \frac{(2x)(\sqrt{x+2}) - (x^2)(\frac{1}{2\sqrt{x+2}})}{(\sqrt{x+2})^2} - \frac{(1/2\sqrt{x+2})(x^2) - (\sqrt{x+2})(2x)}{(x^2)^2} )

Simplify the expression:

( \frac{2x\sqrt{x+2} - \frac{x^2}{2\sqrt{x+2}}}{x+2} - \frac{\frac{x^2}{2\sqrt{x+2}} - 2x\sqrt{x+2}}{x^4} )

( \frac{4x\sqrt{x+2} - x^2\sqrt{x+2} - x^2\sqrt{x+2} + 4x\sqrt{x+2}}{2x^4} )

( \frac{8x\sqrt{x+2} - 2x^2\sqrt{x+2}}{2x^4} )

( \frac{4x\sqrt{x+2} - x^2\sqrt{x+2}}{x^4} )

So, the derivative of ( \frac{x^2}{\sqrt{x+2}} - \frac{\sqrt{x+2}}{x^2} ) is ( \frac{4x\sqrt{x+2} - x^2\sqrt{x+2}}{x^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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