How would I calculate the derivative of f(x)= (x^3+5)/x

Question

anonymous · Answer

Please help me step by step (and explain why where you can)
$$f(x)= \frac{ x ^{3}+5 }{ x }$$

I have the answer, but I can't figure out how to get there: 
$$\frac{ 2x-5 }{ x ^{2} }$$

Note: I'm supposed to be using the power rule.

amistre64 · Answer

if you dont want to implement the usually convoluted quotient rule on a quotient; you might want to up the denominator and run teh product rule

amistre64 · Answer

[(x^3+5) (x^-1) ]' = (x^3+5)' (x)+(x^3+5) (x^-1)'

(3x^2) (x) - (x^3+5) (x^-2)

anonymous · Answer

I have to use the power rule on this one (as per the course requirements.) I'm just not sure how to go about it.

amistre64 · Answer

you have to use more than a power rule .... you have a rational function, a quotient.  you cant just use one single "rule"

amistre64 · Answer

the "power rule" is required within it yes; but you cant just run that rule

anonymous · Answer

I'm allowed to pull the x up to the top temporarily, but I have to use the power rule and then I can bring the x back down.

amistre64 · Answer

$$f(x)= \frac{ x ^{3}+5 }{ x }$$
$$f'(x)= \frac{ x(x ^{3}+5)'-x'(x ^{3}+5) }{ x^2 }$$

amistre64 · Answer

$$f(x)= \frac{ x ^{3}+5 }{ x }$$
$$f(x)= (x ^{3}+5)~(x^{-1})$$
$$f'(x)= (x ^{3}+5)'~(x^{-1})+(x ^{3}+5)~(x^{-1})'$$

amistre64 · Answer

maybe expand it after pulling it up?

$$f(x)= (x ^{3}+5)~(x^{-1})$$
$$f(x)= x ^{2}+5x^{-1}$$
$$f'(x)= 2x -5x^{-2}$$

amistre64 · Answer

if so then its just playing with fractions again
$$2x -5x^{-2}$$
$$2x -\frac{5}{x^{2}}$$
$$\frac{2xx^2}{x^2} -\frac{5}{x^{2}}$$
$$\frac{2x^3-5}{x^2}$$