# How do you find the derivative of y =sqrt(9-x)?

Aug 31, 2014

First you will rewrite the equation to $y = {\left(9 - x\right)}^{\frac{1}{2}}$.

Next you follow the chain rule using the power rule, the constant rule, giving you:

$y ' = \frac{1}{2} {\left(9 - x\right)}^{- \frac{1}{2}} \cdot - 1$

Finally you simplify to:

$y ' = - \frac{1}{2} {\left(9 - x\right)}^{- \frac{1}{2}}$ or $y ' = - \frac{1}{2 \sqrt{9 - x}}$

Normally, when you are finding a derivative, you always change roots to their exponent forms first to simply the process of taking the derivative.

With one function is nested inside the other the chain rule needs to be applied which states:

$F \left(x\right) = f \left(g \left(x\right)\right)$ then $F ' \left(x\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

The power rule for taking a derivative is pretty simple. If:

$f \left(x\right) = {x}^{n}$ then $f ' \left(x\right) = n \cdot {x}^{n - 1}$. $f \left(x\right) = y$