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Neko [114]
4 years ago
10

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

Mathematics
1 answer:
UkoKoshka [18]4 years ago
7 0

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 replaceu = \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

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