How to find this derivative using difference quotient?

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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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sarah desien
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sarah desien

Updated on July 30, 2022

Comments

  • sarah desien
    sarah desien 3 months

    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

  • sarah desien
    sarah desien about 8 years
    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 ^?
  • mookid
    mookid about 8 years
    using difference quotient is painful here. Do you have to do it absolutely this way?
  • sarah desien
    sarah desien about 8 years
    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!
  • mookid
    mookid about 8 years
    this is the way to go.
  • sarah desien
    sarah desien about 8 years
    thank you! this definetly helps!
  • sarah desien
    sarah desien about 8 years
    so 12x^5 is 60x^4
  • mookid
    mookid about 8 years
    exactly.${{{{}}}}$