Question

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

Answer 1

`dy/dx = 3/(2(x-1))`

Explanation:

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

If `y=f(x)` then `f'(x)=dy/dx=dy/(du)(du)/dx`

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

`dy/dx = dy/(dv)(dv)/(du)(du)/dx` etc, or `(dy/dx = dy/color(red)cancel(dv)color(red)cancel(dv)/color(blue)cancel(du)color(blue)cancel(du)/dx)`

So with `y = ln(1-x)^(3/2) => y=3/2ln(1-x)`, Then:

`{ ("Let "u=1-x, => , (du)/dx=-1), ("Then "y=3/2lnu, =>, dy/(du)=3/2*1/u=3/(2u) ) :}`

Using `dy/dx=(dy/(du))((du)/dx)` we get:

`dy/dx = (3/(2u))(-1)`
`:. dy/dx = -3/(2(1-x))`
`:. dy/dx = 3/(2(x-1))`