Question

What is the derivative of `f(x)=(x-1)^2/ln(1/x)^2`?

Answer 1

For the big picture, it's quotient rule. For both numerator and denominator, we'll need separate chain rules.

Explanation:

• Quotient rule: be `y=f(x)/g(x)`, then `y'=(f'g-fg')/g^2`

• Chain rule: `(dy)/(dx)=(dy)/(du)(du)/(dx)`

Let's find the derivatives of both numerator and denominator separately then aggregate them up afterwards.

• Numerator: let's rename `u=x-1`
  `(deltay)/(deltax)=2u(1)=2x-2`

• Denominator: let's rename `u=1/x` and `v=ln(u)`
  `(deltay)/(deltax)=-(2v)/(ux)=-(2ln(1/x))/((1/x)(x^2))=-(2ln(1/x))/x`

Now, let's derivate the whole quotient:

`(dy)/(dx)=((2x-2)(ln(1/x)^2)+(x-1)^2(2ln(1/x))/x)/ln(1/x)^4`

`(dy)/(dx)=(((2x)(x-1)(ln(1/x)^2)+2(x-1)^2ln(1/x))/x)/ln(1/x)^4`

`(dy)/(dx)=((x-1)(ln(1/x))(2xln(1/x)+2(x-1)))/(xln(1/x)^4)`

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