I have to find the derivative of: $$y=\frac{(5x^6-1)}{x^2}$$

I keep on getting this problem wrong. Should I use the quotient rule

$$\frac{f(x)g'(x) - g(x)f'(x)}{g(x)^2}$$ However, my answer is:

$$y'= 30x^2 +\frac{6}{x^4} - 120{x^3}$$

Am I utilizing the wrong method, or have I just evaluated the problem incorrectly?

  • 6
    $\begingroup$ Perhaps manipulating the fuction a bit helps: $$y = \frac{5x^6-1}{x^2} = \frac{5x^6}{x^2}-\frac{1}{x^2} = 5x^4 -x^{-2}$$ $\endgroup$
    – Darth Geek
    Aug 13 '14 at 18:26
  • 1
    $\begingroup$ For this to be understandable, you have to format this using MathJax. Since you have over 400 rep on this site, I assume you know how to do this. By the way, the quotient rule you gave is incorrect by a sign. $\endgroup$
    – rogerl
    Aug 13 '14 at 18:26
  • $\begingroup$ Thanks Dark Greek. I see it now. $\endgroup$
    – Cetshwayo
    Aug 13 '14 at 18:29

$$\frac{dy}{dx}=\frac{(5x^6-1)'x^2-(5x^6-1) (x^2)'}{x^4}=\frac{30x^5 \cdot x^2-2x(5x^6-1)}{x^4}=\frac{30x^7-10x^7+2x}{x^4}=\frac{20x^7+2x}{x^4}=20x^3+\frac{2}{x^3}$$


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