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LilySwan Group Title

find dy/dx if (ax+cx^5)/kx^8

  • 5 months ago
  • 5 months ago

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  1. myininaya Group Title
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    a,c, and k are constants?

    • 5 months ago
  2. LilySwan Group Title
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    yes

    • 5 months ago
  3. micahwood50 Group Title
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    Just use quotient rule.

    • 5 months ago
  4. LilySwan Group Title
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    got stucked at one point

    • 5 months ago
  5. micahwood50 Group Title
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    \(\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}\)

    • 5 months ago
  6. LilySwan Group Title
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    why can we solve the numerator part first

    • 5 months ago
  7. micahwood50 Group Title
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    treat constant as if they are just "number", and not variable.

    • 5 months ago
  8. myininaya Group Title
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    @LilySwan Did you have a problem with finding derivative of the top or the bottom? That is what you need to plug into @micahwood50 's formula. what do you mean solve the numerator? You shouldn't be solving. You should be differentiating it. What is the derivative of the top part of your fraction?

    • 5 months ago
  9. LilySwan Group Title
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    for the upper part 5cx^4 +a

    • 5 months ago
  10. micahwood50 Group Title
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    Right f'(x) = a+5cx^4

    • 5 months ago
  11. micahwood50 Group Title
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    Can you find g'(x)?

    • 5 months ago
  12. LilySwan Group Title
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    for the lower part it will be 8kx^7

    • 5 months ago
  13. micahwood50 Group Title
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    Yep. So you have f(x) = ax+cx^5, f'(x) = a+5cx^4, g(x) = kx^8, g'(x) = 8kx^7 Just plug these in formula given above then simplify.

    • 5 months ago
  14. micahwood50 Group Title
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    Can you do it? @LilySwan

    • 5 months ago
  15. LilySwan Group Title
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    trying wait

    • 5 months ago
  16. LilySwan Group Title
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    i was told to use this formula \[\frac{ d }{ dx }(\frac{ u}{ v }) =\frac{ (v \frac{ du }{ dx })-(u \frac{ dv }{ dx }) }{ v^2 }\]

    • 5 months ago
  17. myininaya Group Title
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    same formula different notation

    • 5 months ago
  18. LilySwan Group Title
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    ?

    • 5 months ago
  19. myininaya Group Title
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    \[\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \] if you want you can replace the f's with u's the g's with v's the ' things with d/dx and you have the same formula

    • 5 months ago
  20. LilySwan Group Title
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    \[5cx^4+a.kx^8 - ax+cx^5. 8kx^7/(kx^8)^2\]

    • 5 months ago
  21. myininaya Group Title
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    \[\frac{(5cx^4+a)(kx^8)-(ax+cx^5)(8kx^7)}{(kx^8)^2}\] i think this is what you mean

    • 5 months ago
  22. LilySwan Group Title
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    yes

    • 5 months ago
  23. LilySwan Group Title
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    what should i be doing the next

    • 5 months ago
  24. myininaya Group Title
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    guess you could multiply things out on top and also use law of exponents on the bottom to simplify the bottom

    • 5 months ago
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