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

1 Answer
Dec 30, 2015

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)#