How do you find the second derivative of # f(x) = x/(1-ln(x-1))# ?

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Answer 1 · 2017-10-10T13:01:12

# f'(x)={2x-1-(x-1)ln(x-1)}/[(x-1){1-ln(x-1)}^2].#

Let us write,

#f(x)=g(x)/(h(x))," where, "g(x)=x, and, h(x)=1-ln(x-1).#

Using the Quotient Rule for Diffn., we have,

#f'(x)={h(x)g'(x)-g(x)h'(x)}/((h(x))^2.#

Now, #h(x)=1-ln(x-1) rArr h'(x)=0-1/(x-1)*d/dx(x-1)#

#:. h'(x)=-1/(x-1).#

Also, #g(x)=x rArr g'(x)=1.#

Utilising these, we get,

#f'(x)={(1-ln(x-1))*1-x(-1/(x-1))}/(1-ln(x-1))^2, i.e., #

#f'(x)={1-ln(x-1)+x/(x-1)}/{1-ln(x-1)}^2, or, #

# f'(x)={2x-1-(x-1)ln(x-1)}/[(x-1){1-ln(x-1)}^2].#