What is the derivative of #y = log3 [(x+1/x-1)^ln3]#?

Question

Emma Aldridge · Accepted Answer

I will guess that this should be: #y = log_3 [(x+1/x-1)^ln3]#, or #y = log_3 [((x+1)/(x-1))^ln3]#.

#y = log_3 [(x+1/x-1)^ln3]# #y =ln3 log_3 (x+1/x-1)# (power property of logarithms) #y =ln3 (ln (x+1/x-1))/ln3# (change of base) #y = ln (x+1/x-1)# So we get: #y' = 1/(x+1/x-1) * d/dx(x+1/x-1)# # = (1-1/x^2)/(x+1/x-1)# # = (x^2-1)/(x^2-x+1)# Note if the original should have been #y = log_3 [((x+1)/(x-1))^ln3]#, Then we get #y = ln((x+1)/(x-1)) = ln(x+1) -ln(x-1)# and #y' = 1/(x+1)-1/(x-1) = (-2)/(x^2-1)#

Elijah Abram · Answer

undefined To find the derivative of $ y = \log_3 \left[ \left( \frac{x+1}{x-1} \right)^{\ln 3} \right] $, we can use the chain rule and the properties of logarithms. The derivative is:

$$ \frac{dy}{dx} = \frac{1}{\ln 3} \cdot \frac{1}{\ln 3} \cdot \frac{x-1}{x+1} \cdot \left( \frac{x+1}{x-1} \right)^{\ln 3} \cdot \left( \frac{x-1}{x+1} \right) $$

$$ = \frac{(x-1)^2}{(x+1)^2} \cdot \left( \frac{x+1}{x-1} \right)^{\ln 3} $$

$$ = \frac{(x-1)^2}{(x+1)^2} \cdot 3^{\ln \left( \frac{x+1}{x-1} \right)} $$

$$ = \frac{(x-1)^2}{(x+1)^2} \cdot \left( \frac{x+1}{x-1} \right)^{\ln 3} $$

$$ = \frac{(x-1)^2 \cdot 3^{\ln \left( \frac{x+1}{x-1} \right)}}{(x+1)^2} $$