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LilySwan

  • 6 months ago

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

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  1. myininaya
    • 6 months ago
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    a,c, and k are constants?

  2. LilySwan
    • 6 months ago
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    yes

  3. micahwood50
    • 6 months ago
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    Just use quotient rule.

  4. LilySwan
    • 6 months ago
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    got stucked at one point

  5. micahwood50
    • 6 months 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. LilySwan
    • 6 months ago
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    why can we solve the numerator part first

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

  8. myininaya
    • 6 months 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. LilySwan
    • 6 months ago
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    for the upper part 5cx^4 +a

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

  11. micahwood50
    • 6 months ago
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    Can you find g'(x)?

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

  13. micahwood50
    • 6 months 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. micahwood50
    • 6 months ago
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    Can you do it? @LilySwan

  15. LilySwan
    • 6 months ago
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    trying wait

  16. LilySwan
    • 6 months 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
    • 6 months ago
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    same formula different notation

  18. LilySwan
    • 6 months ago
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    ?

  19. myininaya
    • 6 months 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. LilySwan
    • 6 months ago
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    \[5cx^4+a.kx^8 - ax+cx^5. 8kx^7/(kx^8)^2\]

  21. myininaya
    • 6 months 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. LilySwan
    • 6 months ago
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    yes

  23. LilySwan
    • 6 months ago
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    what should i be doing the next

  24. myininaya
    • 6 months 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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