# How do you find the derivative (quotient rule) for (x^2 + 8x + 3)/sqrtx?

Aug 15, 2015

${y}^{'} = \frac{1}{2} \cdot {x}^{- \frac{3}{2}} \cdot \left(3 {x}^{2} + 8 x - 3\right)$

#### Explanation:

The quotient rule tells you that you can differentiate functions that can be written as the quotient of two other functions, let's say $f \left(x\right)$ and $g \left(x\right)$ by using the formula

color(blue)(d/dx(y) = ([d/dx(f(x))] * g(x) - f(x) * d/dx(g(x)))/(g(x))^2, with $g \left(x\right) \ne 0$

In your case, you have $f \left(x\right) = {x}^{2} + 8 x + 3$ and $g \left(x\right) = \sqrt{x}$. The derivative of $y$ will thus be

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{\left[\frac{d}{\mathrm{dx}} \left({x}^{2} + 8 x + 3\right)\right] \cdot \sqrt{x} - \left({x}^{2} + 8 x + 3\right) \cdot \frac{d}{\mathrm{dx}} \left(\sqrt{x}\right)}{\sqrt{x}} ^ 2$

${y}^{'} = \frac{\left(2 x + 8\right) \cdot \sqrt{x} - \left({x}^{2} + 8 x + 3\right) \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{x}}}{x}$

${y}^{'} = \frac{\left(2 x + 8\right) \cdot {x}^{\frac{1}{2}} - \frac{1}{2} \cdot \left({x}^{2} + 8 x + 3\right) \cdot {x}^{- \frac{1}{2}}}{x}$

This is equivalent to

${y}^{'} = \frac{{x}^{- \frac{1}{2}} \cdot \left[\left(2 x + 8\right) \cdot x - \frac{1}{2} \left({x}^{2} + 8 x + 3\right)\right]}{x}$

${y}^{'} = \frac{{x}^{- \frac{1}{2}} \cdot \left(2 {x}^{2} + 8 x - \frac{1}{2} {x}^{2} - 4 x - \frac{3}{2}\right)}{x}$

${y}^{'} = \frac{1}{2} \cdot {x}^{- \frac{1}{2}} / x \cdot \left(4 {x}^{2} + 16 x - {x}^{2} - 8 x - 3\right)$

${y}^{'} = \textcolor{g r e e n}{\frac{1}{2} \cdot {x}^{- \frac{3}{2}} \cdot \left(3 {x}^{2} + 8 x - 3\right)}$