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Quickly Check my Work Please|Finding Derivatives

  • Thread starter Raza
  • Start date
  • #1
203
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Homework Statement


[tex]y=\frac{8x^4-5x^2-2}{4x^3}[/tex]

Find y'.

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

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

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

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


[tex]y'=\frac{4x^2(8x^4+5x^2+6)}{4x^2(4x^4)}[/tex]
The 4x2 gets cancelled out.

Finally,

[tex]y'=\frac{8x^4+5x^2+6}{4x^4}[/tex]



Did I do it right?
 
Last edited:

Answers and Replies

  • #2
2,063
2
Perrrfect.

edit: The second and third terms in the numerator (of the final result) needs some corrections.
 
Last edited:
  • #3
203
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Thank you very much!
 
  • #4
286
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Rather than using the quotient rule, it would be quicker to simplify the original expression...

[tex]y=\frac{8x^4-5x^2-2}{4x^3}[/tex]

[tex]= 2x - \frac{5}{4} x^{-1} - \frac{1}{2}x^{-3}[/tex]

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:

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