Question

Differentiating a Logarithmic Function In Exercise, find the derivative of the function.

Answer 1

Answer:

$\frac{d}{dx} ln(\frac{x(x-1)}{x-2}) = \frac{x-2}{x(x-1)} \frac{x^2 -4x +2}{(x-2)^2} = \frac{x^2 -4x +2}{x(x-1)(x-2)}$

Step-by-step explanation:

For this case we want to find the derivate of this function:

$y = ln(\frac{x(x-1)}{x-2})$

And in order to find the derivate we need to apply the chain rule given by:

$\frac{df(u)}{dx} =\frac{df}{du} \frac{du}{dx}$

And on this case $f = ln(u), u = \frac{x(x-1)}{x-2}$

And we can find the partial derivates like this:

$\frac{d}{du} (ln(u))= \frac{1}{u}$

$\frac{d}{dx}(\frac{x(x-1)}{x-2})= \frac{(2x-1)(x-2) -x(x-1)}{(x-2)^2}= \frac{x^2 -4x +2}{(x-2)^2}$

And if we replace we got:

$\frac{d}{dx} ln(\frac{x(x-1)}{x-2}) = \frac{1}{u} \frac{x^2 -4x +2}{(x-2)^2}$

And if we replace$u = \frac{x(x-1)}{x-2}$ we got:

$\frac{d}{dx} ln(\frac{x(x-1)}{x-2}) = \frac{x-2}{x(x-1)} \frac{x^2 -4x +2}{(x-2)^2} = \frac{x^2 -4x +2}{x(x-1)(x-2)}$

And that would be our final answer on this case