ln Problem

[Jason76]

\(\displaystyle f(x) = \dfrac{5}{\ln(x)}\)

\(\displaystyle f'(x) = \dfrac{\ln x (0) - 5 (\dfrac{1}{x})}{[\ln(x)]^{2}}\) - using quotient rule \(\displaystyle \dfrac{g(f') - f(g')}{g^{2}}\) given \(\displaystyle \dfrac{f}{g}\)

\(\displaystyle f'(x) = \dfrac{0 - \dfrac{5}{x}}{[\ln(x)]^{2}}\)

 

[Deleted member 4993]

\(\displaystyle f(x) = \dfrac{5}{\ln(x)}\)

\(\displaystyle f'(x) = \dfrac{\ln x (0) - 5 (\dfrac{1}{x})}{[\ln(x)]^{2}}\) - using quotient rule \(\displaystyle \dfrac{g(f') - f(g')}{g^{2}}\) given \(\displaystyle \dfrac{f}{g}\)

\(\displaystyle f'(x) = \dfrac{0 - \dfrac{5}{x}}{[\ln(x)]^{2}}\)

\(\displaystyle y = 5 * [ln(x)]^{-1}\)

\(\displaystyle y' = 5 * (-1) * \frac{1}{x} * [ln(x)]^{-2}\)

\(\displaystyle y' = \ - \ \dfrac{5}{x * [ln(x)]^{2}}\)

 

[lookagain]

\(\displaystyle f(x) = \dfrac{5}{\ln(x)}\)

\(\displaystyle f'(x) = \dfrac{\ln x (0) - 5 (\dfrac{1}{x})}{[\ln(x)]^{2}}\) - using quotient rule \(\displaystyle \dfrac{g(f') - f(g')}{g^{2}}\) given \(\displaystyle \dfrac{f}{g}\)

\(\displaystyle f'(x) = \dfrac{0 - \dfrac{5}{x}}{[\ln(x)]^{2}}\)

Or, Jason76, continue from your last step:


Multiply the numerator and the denominator by x:


\(\displaystyle f'(x) \ = \ \dfrac{\bigg(\dfrac{x}{1}\bigg)\bigg(\dfrac{-5}{x}\bigg)}{(x)[\ln(x)]^2}\)


\(\displaystyle f'(x) \ = \ \dfrac{-5}{x[\ln(x)]^2}\)

 

[Jason76]

Or, Jason76, continue from your last step:


Multiply the numerator and the denominator by x:


\(\displaystyle f'(x) \ = \ \dfrac{\bigg(\dfrac{x}{1}\bigg)\bigg(\dfrac{-5}{x}\bigg)}{(x)[\ln(x)]^2}\)


\(\displaystyle f'(x) \ = \ \dfrac{-5}{x[\ln(x)]^2}\)

You and Khan got slightly different answers. Who is right?

Oh wait, never mind, I see they're the same.