# How do you differentiate g(y) =sqrtx (x^2 - 2)^(3/2  using the product rule?

Mar 4, 2017

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

#### Explanation:

I'm assuming that "$g \left(y\right)$" in the question is supposed to be "$g \left(x\right)$"
This exercise will require the product rule as well as the chain rule.

Recall that the product rule differentiates two functions that are products of one another.

For the function $h \left(x\right) = f \left(x\right) \cdot g \left(x\right)$,

(1) $h ' \left(x\right) = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$

Recall that the chain rule differentiates a function that is a composition of two functions.

For the function $h \left(x\right) = f \left(g \left(x\right)\right)$

(2) $h ' \left(x\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

To begin, write all powers as exponents:

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

Differentiate

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

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

From here, we clean it up.

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

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