Question

How do you find the derivative for `(x^2 + 3) / x`?

Answer 1

I found: `(x^2-3)/x^2`

Explanation:

You can use the Quotient Rule where the derivative of `f(x)/g(x)` is:
`(f'(x)g(x)-f(x)g'(x))/[g(x)]^2`
in your case you get:

`(2x*x-(x^2+3)*1)/x^2=(2x^2-x^2-3)/x^2=(x^2-3)/x^2`

Answer 2

Alternatively, you could differentiate this function by using the product rule.

Explanation:

You could rewrite your original function as

`y = (x^2 + 3) * x^(-1`

This will allow you to use the product rule, which tells you that functions that can be expressed as the product of two other functions

`y = f(x) * g(x)`

can be differentiate by

`color(blue)(d/dx(y) = f^'(x)g(x) + f(x)g^'(x)`

In your case, you have

`f(x) = x^2 + 3` and `g(x) = x^(-1)`, so that your derivative will be

`d/dx(y) = [d/dx(x^2 + 3)] * x^(-1) + (x^2 + 3) * d/dx(x^(-1))`

`d/dx(y) = 2 * color(red)(cancel(color(black)(x))) * 1/color(red)(cancel(color(black)(x))) + (x^2 + 3) * (-x^(-2))`

`d/dx(y) = 2 - (x^2 + 3)/x^2`

`d/dx(y) = 2 - color(red)(cancel(color(black)(x^2)))/color(red)(cancel(color(black)(x^2))) - 3/x^2 = color(green)(1 - 3/x^2)`

Answer 3

Alternatively again, you could use the power rule.

Explanation:

`f(x)=(x^2+3)/(x)`

`=x^2/x+3/x`

`=x+3/x`

`=x+3x^(-1)`

`f'(x)=1-3x^(-2)`

`f'(x)=1-3/x^(2)`