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how would i find the derivative of $x^8+12x^5-4x^4+10x^3-6x+5$? I know the answer is $8x^7+60x^4-16x^3+30x^2-6$. but how should i solve it using difference quotient, can someone please show the step by step procedure? thank you so much! and by the way this is not my homework question. thanks again

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3 Answers 3

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The steps:

  • threat the sum term after term. Here you have 6 terms. This is allowed by the formula $(u+v)' = u' + v'$.

  • use the formula $(au)' = au'$ when $a$ is a constant.

  • for the various terms, the formula to apply is always the same: $ \frac{d x^n}{dx} = n x^{n-1} $

I think you can do the rest by yourself. Comment if you have any difficulty.


Using difference quotient: I do it only with the $12x^5 $ part, but it is the same with the whole thing. $$\begin{align} 12(x+h)^5 - 12x^5 &= 12(x^5 + 5x^4 h + 10 x^3h^2 + 10 x^2h^3 + 5xh^4 + h^5) - 12x^5 \\&= 60x^4h + 12(10 x^3h^2 + 10 x^2h^3 + 5xh^4 + h^5) \end{align}$$ hence $$ \frac1h \left[12(x+h)^5 - 12x^5 \right] = 60x^4 + h 12(10 x^3h + 10 x^2h^2 + 5xh^3 + h^4) \\ \to_{h\to 0} 60x^4 $$

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  • $\begingroup$ i really have a difficulty doing this question. can you please show me how you would do it using difference quotient? thank you so much ! btw how would i do it by difference quotient not by using that function ^? $\endgroup$ Oct 14, 2014 at 23:23
  • $\begingroup$ using difference quotient is painful here. Do you have to do it absolutely this way? $\endgroup$
    – mookid
    Oct 14, 2014 at 23:25
  • $\begingroup$ yes, i understand theres a lot of work involved in this, which is probably unnecessary. but I really want to now how to do it. Thank you for editing my exponents! $\endgroup$ Oct 14, 2014 at 23:27
  • $\begingroup$ this is the way to go. $\endgroup$
    – mookid
    Oct 14, 2014 at 23:30
  • $\begingroup$ so 12x^5 is 60x^4 $\endgroup$ Oct 14, 2014 at 23:40
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The differentiation operator is linear so all you need to show is the first largest term of the function differentiates to the largest term in the derivative and so on with the rest of the terms. Don't plug the entire function into a limit - do it term by term.

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We have

\begin{align*} f'(x) & = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \\ & = \lim_{h \rightarrow 0} \left[\frac{(x+h)^8 + 12(x+h)^5 - 4(x+h)^4 +10 (x+h)^3 - 6(x+h) + 5}{h} - \frac{(x^8 + 12x^5 - 4x^4 +10 x^3 - 6x - 5)}{h} \right] \end{align*}

Now note that using Newton's binomial formula, we have $(x+h)^n = x^n + nx^{n-1}h + \mathcal{O}(h^2)$ (you may forget about all but the first two terms in the limit). Substitution will yield the desired result.

Alternatively, you can show using the same procedure that $\frac{\mathrm{d}}{\mathrm{d} x} x^n = n x^{n-1}$ and that taking the derivative is a linear operation.

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  • $\begingroup$ thank you! this definetly helps! $\endgroup$ Oct 14, 2014 at 23:31

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