# How do you find the derivative of #(sqrt(x+13))/((x-4)sqrt(2x+1))#?

Use the quotient and product rules.

So, we get:

That's it for the calculus. Now we need to do some algebra to write this in a nicer (or at least, less messy) form.

We'll clear the fractions in the numerator by multiplying the numerator and the denominator by:

We get:

(I think that's correct)

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To find the derivative of (\frac{\sqrt{x+13}}{(x-4)\sqrt{2x+1}}), you can use the quotient rule. Let (u = \sqrt{x+13}) and (v = (x-4)\sqrt{2x+1}). Then, the derivative is given by:

[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} ]

Now, compute (\frac{du}{dx}) and (\frac{dv}{dx}), then substitute into the quotient rule formula to find the derivative.

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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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