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

Jul 27, 2015

Use the quotient and product rules.

#### Explanation:

I also use $\frac{d}{\mathrm{dx}} \left(\sqrt{x}\right) = \frac{1}{2 \sqrt{x}}$

so, the chain rule gives us: $\frac{d}{\mathrm{dx}} \left(\sqrt{u}\right) = \frac{1}{2 \sqrt{u}} \frac{\mathrm{du}}{\mathrm{dx}}$

$f \left(x\right) = \frac{\sqrt{x + 13}}{\left(x - 4\right) \sqrt{2 x + 1}}$

For $f \left(x\right) = \frac{T}{B}$, the quotient rule gives us: $f ' \left(x\right) = \frac{T ' B - T B '}{{B}^{2}}$

To find the derivative of the denominator, we'll use the product rule:
for $g \left(x\right) = F S$, we have $g ' \left(x\right) = F ' S + F S '$

So, we get:

$f ' \left(x\right) = \frac{\frac{1}{2 \sqrt{x + 13}} \left(\left(x - 4\right) \sqrt{2 x + 1}\right) - \sqrt{x + 13} \left[\left(1\right) \sqrt{2 x + 1} + \left(x - 4\right) \frac{1}{2 \sqrt{2 x + 1}} \cdot 2\right]}{\left(x - 4\right) \sqrt{2 x + 1}} ^ 2$

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.

$\frac{\left[\frac{1}{2 \sqrt{x + 13}} \left(\left(x - 4\right) \sqrt{2 x + 1}\right)\right] - \left[\sqrt{x + 13} \sqrt{2 x + 1}\right] - \left[\frac{\sqrt{x + 13} \left(x - 4\right)}{\sqrt{2 x + 1}}\right]}{{\left(x - 4\right)}^{2} \left(2 x + 1\right)}$

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

$\left(2 \sqrt{x + 13} \sqrt{2 x + 1}\right)$

We get:

$\frac{\left(x - 4\right) \left(2 x + 1\right) - 2 \left(x + 13\right) \left(2 x + 1\right) - 2 \left(x + 13\right) \left(x - 4\right)}{2 \sqrt{x + 13} \sqrt{2 x + 1} {\left(x - 4\right)}^{2} \left(2 x + 1\right)}$

$= \frac{\left(2 {x}^{2} - 7 x - 4\right) - 2 \left(2 {x}^{2} + 27 x + 13\right) - 2 \left({x}^{2} + 9 x - 52\right)}{2 \sqrt{x + 13} \sqrt{2 x + 1} {\left(x - 4\right)}^{2} \left(2 x + 1\right)}$

$= \frac{- 4 {x}^{2} - 79 x + 74}{2 \sqrt{x + 13} \sqrt{2 x + 1} {\left(x - 4\right)}^{2} \left(2 x + 1\right)}$

(I think that's correct)