# How do you differentiate f(x)=1/xsqrt(x-1) using the product rule?

Nov 24, 2015

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

#### Explanation:

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

According to the product rule:

$f ' \left(x\right) = {\left(x - 1\right)}^{\frac{1}{2}} \frac{d}{\mathrm{dx}} \left[{x}^{-} 1\right] + {x}^{-} 1 \frac{d}{\mathrm{dx}} \left[{\left(x - 1\right)}^{\frac{1}{2}}\right]$

We can figure out both the derivatives here:

$\frac{d}{\mathrm{dx}} \left[{x}^{-} 1\right] = - {x}^{-} 2 = - \frac{1}{x} ^ 2$

The next requires a simple use of the chain rule:

$\frac{d}{\mathrm{dx}} \left[{\left(x - 1\right)}^{\frac{1}{2}}\right] = \frac{1}{2} {\left(x - 1\right)}^{- \frac{1}{2}} \stackrel{\text{= 1}}{\overbrace{\frac{d}{\mathrm{dx}} \left[x - 1\right]}} = \frac{1}{2 {\left(x - 1\right)}^{\frac{1}{2}}}$

Plug them back in:

f'(x)=-((x-1)^(1/2))/x^2+1/(2x(x-1)^(1/2

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

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