Derivative of fraction without quotient rule

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SUMMARY

The forum discussion centers on finding the derivative of the function \(\frac{x^3-3x^2+4}{x^2}\) without using the quotient or product rule. The user initially transformed the function to \(x - 3 + 4x^{-2}\) and calculated the derivative as \(1 - 0 - 8x^{-3}\), simplifying it to \(\frac{-7}{x^3}\). However, the correct derivative, as stated in the textbook, is \(\frac{x^3-8}{x^3}\). The user acknowledges an algebraic error despite correctly applying calculus principles.

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  • Knowledge of function transformations, such as rewriting fractions
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The following problem appears in my textbook (before it discusses the quotient or product rule, so those rules cannot be used for the answer):

Find the derivative of the function: \frac{x^3-3x^2+4}{x^2}

I brought the denominator to the top and multiplied it out to get {x-3+4x^-2}[/ltex]. I then took the derivative of that to get {1-0-8x^-3}, which can be simplified to \frac{-7}{x^3}.<br /> <br /> However, in the back of my book, the answer is given as \frac{x^3-8}{x^3}.<br /> <br /> Please enlighten me as to where i went wrong.
 
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Your arithmetic:
1 - \frac{8}{x^3} \neq \frac{-7}{x^3}
 
Thanks! For some reason, i did the calculus right but messed up on the algebra. :redface:
 
It happens. :smile:
 

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