# How do you differentiate (3+2x - x^2)^(1/2)?

Sep 24, 2015

Using the chain rule, the derivative is $\frac{1 - x}{{\left(3 + 2 x - {x}^{2}\right)}^{\frac{1}{2}}}$

#### Explanation:

The expression is a function (a square root) of a function (a polynomial of order two) so the chain rule will prove useful.

This compound function might be considered one function (the inner one) wrapped up in the other function (the outer one).

Under the chain rule, the overall derivative is the derivative of the outer function, keeping the inner function as its argument, times the derivative of the inner function.

The outer function is the square root function.
Denoting this outer function, taking some arbitrary argument $k$, by
$f \left(k\right) = {k}^{\frac{1}{2}}$

this has derivative

$f ' \left(k\right) = \left(\frac{1}{2}\right) {k}^{- \frac{1}{2}}$

in this particular case, $k = \left(3 + 2 x - {x}^{2}\right)$

The inner function is a simple polynomial. Denoting this by

$g \left(x\right) = 3 + 2 x - {x}^{2}$

it might be noted that this has derivative

$g ' \left(x\right) = 2 - 2 x$

The overall derivative is the product of these two derivatives, that is,

$\left(\frac{1}{2}\right) {\left(3 + 2 x - {x}^{2}\right)}^{- \frac{1}{2}} \left(2 - 2 x\right)$

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

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