Studying for a midterm, and one of the problems is: $$\frac{x^3+7}{x}$$ and we have to find the derivative.
My professor is getting: $$2x-\frac{7}{x^2}$$
But I got $$3x-\frac{x^3+7}{x^2}$$
I even tried using an online calculator to verify my results, and I indeed it got the same thing. Can someone tell me what I am doing wrong here? Or am I right?
$$2x-\frac7{x^2}=\frac{2x^3-7}{x^2}$$
$$3x-\frac{x^3+7}{x^2}=\frac{3x^3-(x^3+7)}{x^2}=?$$
If you simplify your answer, you can see that it is in fact the same as your professor's answer. $$3x-\frac{x^3+7}{x^2}=3x-\left(\frac{x^3}{x^2}+\frac 7 {x^2}\right) = 3x-x-\frac 7 {x^2}=2x-\frac 7 {x^2}$$
You and your professor both get the same result: $$3x-\frac{x^3+7}{x^2} = 3x - \left(\frac{x^3}{x^2}+\frac 7{x^2}\right) =$$ $$ = 3x - x - \frac 7{x^2} = 2x - \frac{7}{x^2}$$
