How do you find the derivative of #f(x)=log_2(x^2/(x-1))#?
1 Answer
Feb 18, 2017
Explanation:
We need to know the logarithm rules:
#log_a(b/c)=log_a(b)-log_a(c)# #log_a(b^c)=clog_a(b)# #log_a(b)=log_c(b)/log_c(a)=ln(b)/ln(a)#
We also have to know that:
#d/dx(ln(x))=1/x# #d/dxln(f(x))=1/f(x)*f'(x)#
Then:
#f(x)=log_2(x^2)-log_2(x-1)#
#f(x)=ln(x^2)/ln(2)-ln(x-1)/ln(2)#
#f(x)=1/ln(2)(2ln(x)-ln(x-1))#
The derivative is then:
#f'(x)=1/ln(2)(2(1/x)-1/(x-1)(x-1)')#
#f'(x)=1/ln(2)(2/x-1/(x-1))#
Which can be simplified:
#f'(x)=1/ln(2)((2(x-1)-x)/(x(x-1)))#
#f'(x)=(x-2)/(x(x-1)(ln(2)))#
