How do you find the derivative of $\frac{x^{3} - 3 x^{2} + 4}{x^{2}}$ $(x^3-3x^2+4)/x^2$ ?

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Answer 1 · 2015-06-03T21:25:11

Using the quotient rule, which states that

Be $y = \frac{f (x)}{g (x)}$ $y=f(x)/g(x)$ , $(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/(g(x))^2$

Applying the rule:

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

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

How do you find the derivative of (x^3-3x^2+4)/x^2x3−3x2+4x2(x^3-3x^2+4)/x^2?