Question

How do you find the derivative of `(x^3-3x^2+4)/x^2`?

Answer 1

Using the quotient rule, which states that

Be `y=f(x)/g(x)`, `(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/(g(x))^2`

Applying the rule:

`(dy)/(dx)=((3x^2-6x)x^cancel(2)-(x^3-3x^2+4)(2cancel(x)))/x^(cancel(4)3)`

`(dy)/(dx)=(3x^3cancel(-6x^2)-2x^3cancel(+6x^2)-8)/x^3=(x^3+8)/x^3`