# Question 4ab09

Aug 15, 2017

$\frac{1}{4}$

#### Explanation:

The slope of a function's tangent line at a point is given by the function's derivative. So, we need to find the derivative of $\sqrt{x - 5} = {\left(x - 5\right)}^{\frac{1}{2}}$.

Note that $\frac{d}{\mathrm{dx}} {x}^{\frac{1}{2}} = \frac{1}{2} {x}^{- \frac{1}{2}}$. So, the chain rule states that if we have a function to the $1 / 2$ power instead, we follow the same process of the power rule, by moving down the $1 / 2$ as a multiplicative constant and reducing the power by $1$ (here, to get to -1//2), but we also multiply by the derivative of the inner function.

Thus, $\frac{d}{\mathrm{dx}} {\left(f \left(x\right)\right)}^{\frac{1}{2}} = \frac{1}{2} {\left(f \left(x\right)\right)}^{- \frac{1}{2}} \cdot f ' \left(x\right)$.

So, the derivative of ${\left(x - 5\right)}^{\frac{1}{2}}$ is:

$\frac{d}{\mathrm{dx}} {\left(x - 5\right)}^{\frac{1}{2}} = \frac{1}{2} {\left(x - 5\right)}^{- \frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} \left(x - 5\right)$

The derivative of $x - 5$ is just $1$, so:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {\left(x - 5\right)}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x - 5}}$

The slope of the tangent line to the original function $y$ is found by plugging $x = 9$ into $\mathrm{dy} / \mathrm{dx}$:

1/(2sqrt(9-5))=color(red)(1/4#