# How do you differentiate y= ln(1-x)^(3/2)?

Nov 9, 2016

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

#### Explanation:

If you are studying maths, then you should learn the Chain Rule for Differentiation, and practice how to use it:

If $y = f \left(x\right)$ then $f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

I was taught to remember that the differential can be treated like a fraction and that the "$\mathrm{dx}$'s" of a common variable will "cancel" (It is important to realise that $\frac{\mathrm{dy}}{\mathrm{dx}}$ isn't a fraction but an operator that acts on a function, there is no such thing as "$\mathrm{dx}$" or "$\mathrm{dy}$" on its own!). The chain rule can also be expanded to further variables that "cancel", E.g.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$ etc, or $\left(\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\textcolor{red}{\cancel{\mathrm{dv}}}} \frac{\textcolor{red}{\cancel{\mathrm{dv}}}}{\textcolor{b l u e}{\cancel{\mathrm{du}}}} \frac{\textcolor{b l u e}{\cancel{\mathrm{du}}}}{\mathrm{dx}}\right)$

So with $y = \ln {\left(1 - x\right)}^{\frac{3}{2}} \implies y = \frac{3}{2} \ln \left(1 - x\right)$, Then:

$\left\{\begin{matrix}\text{Let "u=1-x & => & (du)/dx=-1 \\ "Then } y = \frac{3}{2} \ln u & \implies & \frac{\mathrm{dy}}{\mathrm{du}} = \frac{3}{2} \cdot \frac{1}{u} = \frac{3}{2 u}\end{matrix}\right.$

Using $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{\mathrm{dy}}{\mathrm{du}}\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$ we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{3}{2 u}\right) \left(- 1\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3}{2 \left(1 - x\right)}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{2 \left(x - 1\right)}$