How do you find the derivative of #h(x)=log_10((x^2-1)/x)#?

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Answer 1 · 2017-10-12T22:10:09

Convert the base 10 logarithm to the natural logarithm and add the domain restriction

#h(x)=1/ln(10)ln((x^2-1)/x); x>1#

Factor the numerator:

#h(x)=1/ln(10)ln(((x-1)(x+1))/x); x>1#

Use the property that multiplication / division within the in argument is addition / subtraction of logarithms.

#h(x)=1/ln(10)(ln(x-1)+ ln(x+1)-ln(x)); x>1#

Differentiate each term:

#h'(x)=1/ln(10)(1/(x-1)+ 1/(x+1)-1/(x)); x > 1#