`y=log_10((x^2-1)/x)`
The derivative formula of a logarithm is
`d/dx[log_a (u)] = 1/(ln(a) * u)* (du)/dx`
Applying this formula, the derivative of the function will be
`(dy)/dx = d/dx[log_10 ((x^2-1)/x)]`
`(dy)/dx =1/(ln (10) * ((x^2-1)/x)) * d/dx ((x^2-1)/x)`
`(dy)/dx = x/((x^2-1)ln(10) ) * d/dx((x^2-1)/x)`
To get the derivative of `(x^2-1)/x`, apply quotient rule `d/dx ((f(x))/(g(x))) = (g(x)*f'(x) - f(x)*g'(x))/[g(x)]^2` .
`(dy)/dx = x/((x^2-1)ln(10) ) * (x*2x - (x^2-1)*1)/x^2`
`(dy)/dx = x/((x^2-1)ln(10)) * (2x^2 - x^2 + 1)/x^2`
`(dy)/(dx) = x/((x^2-1)ln(10)) * (x^2+1)/x^2`
`(dy)/dx = 1/((x^2-1)ln(10)) * (x^2+1)/x`
`(dy)/dx = (x^2+1)/(x(x^2-1)ln(10))`
Therefore, the derivative of the function is `(dy)/dx = (x^2+1)/(x(x^2-1)ln(10))` .
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