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anonymous

  • 2 years ago

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

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  1. myininaya
    • 2 years ago
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    a,c, and k are constants?

  2. anonymous
    • 2 years ago
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    yes

  3. anonymous
    • 2 years ago
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    Just use quotient rule.

  4. anonymous
    • 2 years ago
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    got stucked at one point

  5. anonymous
    • 2 years ago
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    \(\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}\)

  6. anonymous
    • 2 years ago
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    why can we solve the numerator part first

  7. anonymous
    • 2 years ago
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    treat constant as if they are just "number", and not variable.

  8. myininaya
    • 2 years ago
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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?

  9. anonymous
    • 2 years ago
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    for the upper part 5cx^4 +a

  10. anonymous
    • 2 years ago
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    Right f'(x) = a+5cx^4

  11. anonymous
    • 2 years ago
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    Can you find g'(x)?

  12. anonymous
    • 2 years ago
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    for the lower part it will be 8kx^7

  13. anonymous
    • 2 years ago
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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.

  14. anonymous
    • 2 years ago
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    Can you do it? @LilySwan

  15. anonymous
    • 2 years ago
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    trying wait

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

  17. myininaya
    • 2 years ago
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    same formula different notation

  18. anonymous
    • 2 years ago
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    ?

  19. myininaya
    • 2 years ago
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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

  20. anonymous
    • 2 years ago
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    \[5cx^4+a.kx^8 - ax+cx^5. 8kx^7/(kx^8)^2\]

  21. myininaya
    • 2 years ago
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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

  22. anonymous
    • 2 years ago
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    yes

  23. anonymous
    • 2 years ago
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    what should i be doing the next

  24. myininaya
    • 2 years ago
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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

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