What is the derivative of #f(x)=(x-1)^2/ln(1/x)^2#?
1 Answer
For the big picture, it's quotient rule. For both numerator and denominator, we'll need separate chain rules.
Explanation:
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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.
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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: