`y = log_10((x^2-1)/x)` Find the derivative of the function

`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))` .