What is the derivative of f(x) = log base 10 ((x^2 - 1)/x)?

I got ((x^2 + 1)/(x(x^2 - 1)ln10). Is this correct?

2 Answers
Jul 21, 2018

#f'(x)=1/ln10{(x^2+1)/(x(x^2-1))}#

Explanation:

We know that ,
#color(red)((1)ln(M/N)=lnM-lnN#
#color(blue)((2)log_eX=lnX#
Here ,

#f(x)=log_10((x^2-1)/x)to" Use Change of base formula"#

#:.f(x)=log_e((x^2-1)/x)/log_e10to"Where , e is the new base."#

#:.f(x)=1/ln10color(red)({ln(x^2-1)-lnx}to[Apply(1)])#

Diff.w.r.t # x # , we get

#:.f'(x)=1/ln10{1/(x^2-1)*2x-1/x}#

#:.f'(x)=1/ln10{(2x)/(x^2-1)-1/x}#

#:.f'(x)=1/ln10{(2x^2-x^2+1)/(x(x^2-1))}#

#:.f'(x)=1/ln10{(x^2+1)/(x(x^2-1))}#

#f'(x)=1/\ln10\frac{x^2+1}{x(x^2-1)}#

Explanation:

The given function:

#f(x)=\log_{10}(\frac{x^2-1}{x})#

#f(x)=\log_{e}(\frac{x^2-1}{x})/{ \log_{e}(10)}#

#f(x)=1/\ln10\ln(\frac{x^2-1}{x})#

Differentiating above function w.r.t. #x# using chain rule as follows

#d/dxf(x)=d/dx(1/\ln10\ln(\frac{x^2-1}{x}))#

#f'(x)=1/\ln10\frac{1}{(\frac{x^2-1}{x})}d/dx(\frac{x^2-1}{x})#

#=\1/\ln10frac{x}{(x^2-1)}(\frac{xd/dx(x^2-1)-(x^2-1)d/dx(x)}{x^2})#

#=\1/\ln10frac{x}{(x^2-1)}(\frac{x(2x)-(x^2-1)(1)}{x^2})#

#=\1/\ln10frac{x}{(x^2-1)}(\frac{x^2+1}{x^2})#

#=1/\ln10\frac{x^2+1}{x(x^2-1)}#