How do you find the derivative of f(x)=log_2(x^2/(x-1))?

Feb 18, 2017

$f ' \left(x\right) = \frac{x - 2}{x \left(x - 1\right) \left(\ln \left(2\right)\right)}$

Explanation:

We need to know the logarithm rules:

• ${\log}_{a} \left(\frac{b}{c}\right) = {\log}_{a} \left(b\right) - {\log}_{a} \left(c\right)$
• ${\log}_{a} \left({b}^{c}\right) = c {\log}_{a} \left(b\right)$
• ${\log}_{a} \left(b\right) = {\log}_{c} \frac{b}{\log} _ c \left(a\right) = \ln \frac{b}{\ln} \left(a\right)$

We also have to know that:

• $\frac{d}{\mathrm{dx}} \left(\ln \left(x\right)\right) = \frac{1}{x}$
• $\frac{d}{\mathrm{dx}} \ln \left(f \left(x\right)\right) = \frac{1}{f} \left(x\right) \cdot f ' \left(x\right)$

Then:

$f \left(x\right) = {\log}_{2} \left({x}^{2}\right) - {\log}_{2} \left(x - 1\right)$

$f \left(x\right) = \ln \frac{{x}^{2}}{\ln} \left(2\right) - \ln \frac{x - 1}{\ln} \left(2\right)$

$f \left(x\right) = \frac{1}{\ln} \left(2\right) \left(2 \ln \left(x\right) - \ln \left(x - 1\right)\right)$

The derivative is then:

$f ' \left(x\right) = \frac{1}{\ln} \left(2\right) \left(2 \left(\frac{1}{x}\right) - \frac{1}{x - 1} \left(x - 1\right) '\right)$

$f ' \left(x\right) = \frac{1}{\ln} \left(2\right) \left(\frac{2}{x} - \frac{1}{x - 1}\right)$

Which can be simplified:

$f ' \left(x\right) = \frac{1}{\ln} \left(2\right) \left(\frac{2 \left(x - 1\right) - x}{x \left(x - 1\right)}\right)$

$f ' \left(x\right) = \frac{x - 2}{x \left(x - 1\right) \left(\ln \left(2\right)\right)}$