Quickly Check my Work Please|Finding Derivatives

[Raza]

Homework Statement

y=\frac{8x^4-5x^2-2}{4x^3}

Find y'.

2. The attempt at a solution
y=\frac{8x^4-5x^2-2}{4x^3}

y'=\frac{(32x^3-10x)(4x^3)-(12x^2)(8x^4-5x^2-2)}{(4x^3)^2}

y'=\frac{128x^6-40x^4-96x^6+60x^4+24x^2}{16x^6}

y'=\frac{32x^6+20x^4+24x^2}{16x^6}


y'=\frac{4x^2(8x^4+5x^2+6)}{4x^2(4x^4)}
The 4x² gets canceled out.

Finally,

y'=\frac{8x^4+5x^2+6}{4x^4}



Did I do it right?

 

[neutrino]

Perrrfect.

edit: The second and third terms in the numerator (of the final result) needs some corrections.

 

[Raza]

Thank you very much!

 

[drpizza]

Rather than using the quotient rule, it would be quicker to simplify the original expression...

y=\frac{8x^4-5x^2-2}{4x^3}

= 2x - \frac{5}{4} x^{-1} - \frac{1}{2}x^{-3}

Then, take the derivative. The form of the answer won't be the same as what you have, but it'll be equivalent.