# How do you find the derivative of # (3x^2 + 3x + 4) / sqrt(x)#?

The first to step is to rewrite this so that it looks more like a polynomial. Use numerical exponents:

Split up the numerator:

Divide the terms:

Now, differentiate each term individually:

This gives us a derivative of:

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To find the derivative of the function (\frac{3x^2 + 3x + 4}{\sqrt{x}}), you can use the quotient rule. The quotient rule states that for functions (u(x) = \frac{f(x)}{g(x)}), the derivative is given by:

[u'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}]

Now, let (f(x) = 3x^2 + 3x + 4) and (g(x) = \sqrt{x}). Then, (f'(x)) and (g'(x)) can be calculated as follows:

[f'(x) = 6x + 3]

[g'(x) = \frac{1}{2\sqrt{x}}]

Now, plug these values into the quotient rule formula:

[u'(x) = \frac{(6x + 3)\sqrt{x} - (3x^2 + 3x + 4)\left(\frac{1}{2\sqrt{x}}\right)}{(\sqrt{x})^2}]

[u'(x) = \frac{(6x + 3)\sqrt{x} - \frac{3x^2 + 3x + 4}{2\sqrt{x}}}{x}]

[u'(x) = \frac{(6x + 3)\sqrt{x} - \frac{3x^2}{\sqrt{x}} - \frac{3x}{\sqrt{x}} - \frac{4}{\sqrt{x}}}{2\sqrt{x}}]

[u'(x) = \frac{(6x + 3)\sqrt{x} - \frac{3x^{\frac{3}{2}}}{\sqrt{x}} - 3\sqrt{x} - \frac{4}{\sqrt{x}}}{2\sqrt{x}}]

[u'(x) = \frac{(6x + 3)\sqrt{x} - 3x^{\frac{3}{2}} - 3\sqrt{x} - \frac{4}{\sqrt{x}}}{2\sqrt{x}}]

This expression represents the derivative of the given function.

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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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