What is the derivative of #log_2(x^2/(x-1))#?

Question

Paisley Johnson · Accepted Answer

#2/(xln(2))- 1/((x - 1)ln(2)) #

#log_2(x^2/(x - 1)) = #

#log_2(x^2)- log_2(x - 1) = #

#2log_2(x)- log_2(x - 1) = #

#(2/ln(2))ln(x)- 1/ln(2)ln(x - 1) #

Differentiate:

#(2/ln(2))1/(x)- 1/ln(2)1/(x - 1) = #

Simplify:

#2/(xln(2))- 1/((x - 1)ln(2)) #

Elijah Abram · Answer

#=1/ln2((x-2)/(x(x-1)))# Use #log_ba=lna/lnb#. #(log_2(x^2/(x-1))'# #=(ln (x^2/(x-1))/(ln2))'# #=1/ln2(lnx^2-ln(x-1))'# #1/ln2(2(lnx)'-(ln(x-1))')# #=1/ln2(2/x-1/(x-1))# #=1/ln2((x-2)/(x(x-1)))#

Ezekiel Alexander · Answer

undefined To find the derivative of $ \log_2\left(\frac{x^2}{x-1}\right) $, you can use the chain rule and the properties of logarithms. The derivative is:

$$ \frac{d}{dx}\left(\log_2\left(\frac{x^2}{x-1}\right)\right) = \frac{1}{\ln(2)} \cdot \frac{d}{dx}\left(\frac{x^2}{x-1}\right) $$

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

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

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

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

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