Verified answer

The easiest way is to simplify it by using property of logs: log(a/b) = log(a) - log(b)

-->

log₁₀( x / (x - 1) ) = log₁₀(x) - log₁₀(x - 1)

--> convert to natural log, recal log_a(x) = log_b(x) / log_b(a)

log₁₀(x/(x - 1) = { ln(x) - ln(x - 1) } / ln(10)

--> so the derivative is easy now for both (just multiply by 1/ln(10) and subtract)

{ 1/x - 1/(x - 1) } / ln(10)

--> which is

(x - 1 - x) / { x(x - 1)ln(10) } = -1 / (ln(10) * x(x - 1) }

Edit:

If you don't do that simplification first, you still get the same value, but you have to use the chain rule:

log₁₀(x/(x - 1)) = ln(x/(x - 1)) / ln(10)

--> derivative of inside, use quotient rule

{ (x - 1) - x } / (x - 1)² = -1/(x - 1)²

--> now multiply by reciprocal (derivative of natural log):

(x - 1) / x * -1/(x - 1)² = -1 / (x(x - 1))

--> now just divide by the ln(10) and you have the same thing as above

"log base 10 times (x/x-1)" doesn't make sense

1/((x - x^2) Log[10])

First rewrite as log(x)-log(x-1)

differentiate one term at a time

y=logx<->10^y=x

yln10=lnx

y=lnx/ln10

dy/dx=1/xln10

y=log(x-1)

10^y=x-1

yln10=ln(x-1)

y=ln(x-1)/ln10

dy/dx=1/(x-1)ln10

Final Answer

dy/dx=1/xln10-1/(x-1)ln10