# How do you find the derivative for #sqrt(x)/(x^3+1)#?

You can differentiate this function by using the quotient rule, which allows you to differentiate a function that can be written as

by using the formula

In your case, you can say that

This means that you can write

This can be rewritten as

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To find the derivative of ( \frac{\sqrt{x}}{x^3+1} ), you can use the quotient rule. The quotient rule states that if you have a function of the form ( \frac{u}{v} ), then its derivative is given by ( \frac{u'v - uv'}{v^2} ), where ( u' ) and ( v' ) are the derivatives of ( u ) and ( v ) with respect to ( x ), respectively.

Let ( u = \sqrt{x} ) and ( v = x^3 + 1 ).

Then, ( u' = \frac{1}{2\sqrt{x}} ) (by the power rule) and ( v' = 3x^2 ) (by the power rule).

Now, apply the quotient rule:

[ \frac{d}{dx} \left( \frac{\sqrt{x}}{x^3+1} \right) = \frac{\left( \frac{1}{2\sqrt{x}} \right)(x^3 + 1) - (\sqrt{x})(3x^2)}{(x^3 + 1)^2} ]

[ = \frac{(x^3 + 1) - 6x^2\sqrt{x}}{2x\sqrt{x}(x^3 + 1)} ]

[ = \frac{x^3 + 1 - 6x^{\frac{5}{2}}}{2x\sqrt{x}(x^3 + 1)} ]

[ = \frac{x^3 + 1 - 6x^{\frac{5}{2}}}{2x^{\frac{3}{2}}(x^3 + 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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