# Quickly Check my Work Please|Finding Derivatives

1. Feb 15, 2007

### Raza

1. The problem statement, all variables and given/known data
$$y=\frac{8x^4-5x^2-2}{4x^3}$$

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 cancelled out.

Finally,

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

Did I do it right?

Last edited: Feb 15, 2007
2. Feb 15, 2007

### neutrino

Perrrfect.

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

Last edited: Feb 15, 2007
3. Feb 15, 2007

### Raza

Thank you very much!

4. Feb 15, 2007

### 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.

Last edited: Feb 15, 2007