# How do you find the derivative of 1/sqrt (x-1)?

Aug 1, 2017

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

#### Explanation:

We need

$\left({x}^{n}\right) ' = n {x}^{n - 1}$

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

$\forall x \in \left(1 , + \infty\right)$, $f \left(x\right) \in \left(0 , + \infty\right)$

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

Aug 1, 2017

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

#### Explanation:

$\text{express } y = \frac{1}{\sqrt{x - 1}} = {\left(x - 1\right)}^{- \frac{1}{2}}$

$\text{differentiate using the "color(blue)"chain rule}$

$\text{given "y=f(g(x))" then}$

dy/dx=f'(g(x)xxg'(x)larr" chain rule"

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2} {\left(x - 1\right)}^{- \frac{3}{2}} \times \frac{d}{\mathrm{dx}} \left(x - 1\right)$

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = - \frac{1}{2 \sqrt{{\left(x - 1\right)}^{3}}}$