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anonymous

  • 5 years ago

Use the Chain Rule to find dw/dt. (Enter your answer only in terms of t.) w = xey /z, x = t6, y = 9 - t, z = 4 + 4t

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  1. anonymous
    • 5 years ago
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    Do you mean find dw/dt where\[w=\frac{xe^y}{z}, x=t^6, y=9-t, z=4+4t?\]

  2. anonymous
    • 5 years ago
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    yes

  3. anonymous
    • 5 years ago
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    i got an answer I keep on getting the same one, but my hw is online and it marks it wrong

  4. anonymous
    • 5 years ago
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    OK. If you really *must* use the chain rule to solve,\[\frac{dw}{dt}=\frac{dw}{dx}\frac{dx}{dy}\frac{dy}{dz}\frac{dz}{dt}\]Then expand dx/dy and dy/dz again\[\frac{dw}{dt}=\frac{dw}{dx}(\frac{dx}{dt}\frac{dt}{dy})(\frac{dy}{dt}\frac{dt}{dz})\frac{dz}{dt}\]

  5. anonymous
    • 5 years ago
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    this is for calc 3 so I have to multiply the leibniz notation and add them

  6. anonymous
    • 5 years ago
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    Ah, okay, knowing the level helps.

  7. anonymous
    • 5 years ago
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    Well, in that case, it's just\[\frac{dw}{dt}=\frac{dw}{dx}\frac{dx}{dt}+\frac{dw}{dy}\frac{dy}{dt}+\frac{dw}{dz}\frac{dz}{dt}\] where all the 'd's are partial.

  8. anonymous
    • 5 years ago
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    yes I have computed that

  9. anonymous
    • 5 years ago
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    I'm doing the rest...

  10. anonymous
    • 5 years ago
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    hey lokisan do you get the divisible sign to be straight instead of this /

  11. anonymous
    • 5 years ago
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    I get \[\frac{dw}{dt}=\frac{-4t^6}{(4+4t)^2}e^{9-t}\]

  12. anonymous
    • 5 years ago
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    Did you need just a check on your answer, or the full-on working?

  13. anonymous
    • 5 years ago
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    nadeem, you type "frac{}{}" into the equation editor. Your numerator and denominator go in the first and second parentheses respectively.

  14. anonymous
    • 5 years ago
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    that's wassup..... appreciate it

  15. anonymous
    • 5 years ago
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    no probs

  16. anonymous
    • 5 years ago
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    is that what you get? how did you get that?

  17. anonymous
    • 5 years ago
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    Just doing what each derivative asks of me first\[\frac{dw}{dt}=(\frac{e^y}{z}).t^6+\frac{xe^y}{z}.(-1)+(-\frac{xe^y}{z^2}).4\]

  18. anonymous
    • 5 years ago
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    Then substitute each of the x=t^6, y=... into the above. You should find that the first two quotients cancel out, and you're left with the last quotient -> the answer.

  19. anonymous
    • 5 years ago
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    This is what I got: \[\frac{dw}{dt}=\frac{-4t^5e^{9-t}(t^2-4t-6)}{(4+4t)^2}\]

  20. anonymous
    • 5 years ago
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    \[\frac{dw}{dt}=(\frac{e^{9-t}}{4+4t}).t^6-\frac{t^6e^{9-t}}{4+4t}-4\frac{t^6e^{9-t}}{(4+4t)^2}\]

  21. anonymous
    • 5 years ago
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    The first two parts cancel, the third is left over.

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