How do you Differentiate (x^2+1)/x

Thanks

This question is best put in the Mathematics category, but this still works.

To differentiate a fraction with the variable on top and bottom, use the Quotient rule. You will also need the Chain rule and Power rule as well.

Quotient Rule: Derivative of the top times the bottom, minus the derivative of the bottom times the top, divided by the bottom squared.
Power rule: Multiply the exponent to the front and reduce the exponent by one.
Chain rule: Derivative of the outside times the derivative of the inside. Keep going until you run out of insides.

y = (x^2+1) / x
dy/dx = [ 2x * x - 1 (x^2 + 1) ] / x^2
dy/dx = ( 2x^2 - x^2 - 1 ) / x^2
dy/dx = ( x^2 - 1 ) / x^2
dy/dx = 1 - 1/x^2

Another approach would be to expand and simplify first, then take the derivative.
y = (x^2 + 1) / x
y = x^2 / x + 1/x
y = x + x^(-1)
dy/dx = 1 -x^(-2)
This yields the same answer.

Using the General Derivative (lim h->0 [ f(x+h) - f(x) ] / h):
f'(x) = lim h->0 [ f(x+h) - f(x) ] / h
f'(x) = lim h->0 [ ( [x+h] ^2 + 1 ) / (x+h) - ( x^2 + 1) / x ] / h ]
f'(x) = lim h->0 [ (x^2+2hx+h^2 + 1 ) / (x+h) - ( x^2 + 1) / x ] / h
f'(x) = lim h->0 [ (x^2+2hx+h^2 + 1 )x / (x+h)x - (x+h)( x^2 + 1) / x(x+h) ] / h
f'(x) = lim h->0 [ x^3+2hx^2+xh^2 + x - x^3 - x - hx^2 - h ] / h x(x+h)
f'(x) = lim h->0 [ hx^2 + xh^2 - h ] / h x(x+h)
f'(x) = lim h->0 [ x^2 + xh - 1 ] / x(x+h)
f'(x) = ( x^2 + x(0) - 1 ) / x(x+0)
f'(x) = ( x^2 - 1 ) / x^2
f'(x) = 1 - 1 / x^2
The also yields the same answer.

Use product rule of differentiation.
If y = u / v
dy/dx = (v du/dx - u dv/dx) / v²

Take x² + 1 as u and x as v

Then you work out like this:

y = (x² + 1) / x
dy / dx = ( x d/dx(x²+1) - (x²+1) d/dx(x) ) / x²
= (x * 2x - (x² + 1) * 1) / x²
= (2x² - x² - 1) / x²
= (x² - 1) / x² => It should be the answer but I'm not sure because I did it quickly. The rule is correct so just do it yourself with the rule.