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monroe17

  • 2 years ago

help? step by step Find dy/dx

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  1. monroe17
    • 2 years ago
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    \[y=(12\sqrt[3]{x}-\frac{ 1 }{ 4x }+e^2)\]

  2. monroe17
    • 2 years ago
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    I know I should rewrite it out.. 12*x^(1/3) but now I'm stuck..?

  3. abb0t
    • 2 years ago
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    \[\sqrt[3]{x} = x^(\frac{ 1 }{ 3 }\] by the property, you subtract 1. So 1/3 - 1 = ? \[\frac{ 1 }{ 4x } = \frac{ 1 }{ 4 }x^{-1} \] and \[e^2 \] is just a constant.

  4. abb0t
    • 2 years ago
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    \[x^{\frac{ 1 }{ 3 }}\]

  5. monroe17
    • 2 years ago
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    okay, thanks.

  6. abb0t
    • 2 years ago
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    \[y' = \frac{ 4 }{ x^{\frac{ 2 }{ 3 } }} - \frac{ 1 }{ 4x^2 }\]

  7. monroe17
    • 2 years ago
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    so.. rewritten it's \[(12*x^{1/3}-\frac{ 1 }{ 4 }x^{-1}+e^2)\]

  8. abb0t
    • 2 years ago
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    sorry, change the - to +.

  9. monroe17
    • 2 years ago
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    ahhh i messed up on the original equation

  10. monroe17
    • 2 years ago
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    \[(12\sqrt[3]{x}-\frac{ 1 }{ 4x }+e^2)^7\]

  11. monroe17
    • 2 years ago
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    so it's chain rule right?

  12. abb0t
    • 2 years ago
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    It's the same thing. now you apply chain rule.

  13. abb0t
    • 2 years ago
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    \[7(4x^{\frac{ -2 }{ 3 }} + \frac{ 1 }{ 4x^2 })^6\]

  14. monroe17
    • 2 years ago
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    mm.. the answer is different then that.

  15. monroe17
    • 2 years ago
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    \[7(12\sqrt[3]{x}-\frac{ 1 }{ 4x }+e^2)^6(4x^{-2/3}+\frac{ 1 }{4 }x^{-2})\]

  16. abb0t
    • 2 years ago
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    oh yeah, you multiply by the deriv again.

  17. monroe17
    • 2 years ago
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    why do you multiply by the derivative?

  18. monroe17
    • 2 years ago
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    oh nevermind haha

  19. monroe17
    • 2 years ago
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    I got it.

  20. monroe17
    • 2 years ago
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    I'm forgetting my simple rulessss! Ahhhh D;

  21. abb0t
    • 2 years ago
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    chain rule states f'(g(x))*g'(x)

  22. abb0t
    • 2 years ago
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    it's all good, i totally 4got chain rule for a moment. im rusty in my math

  23. monroe17
    • 2 years ago
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    thanks so much though ;D

  24. Outkast3r09
    • 2 years ago
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    i remember the chain rule by ths let u equal some function within a function and y=\[f(u)\] so you have\[y=f(u)\] \[\frac{dy}{dx}=\frac{du}{dx}\frac{dy}{du}\] if you remember one party like du/dx or dy/du.... you'll know what you ned to cancel

  25. Outkast3r09
    • 2 years ago
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    |dw:1355374308512:dw|

  26. Outkast3r09
    • 2 years ago
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    \[(x-2)^x\]

  27. monroe17
    • 2 years ago
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    ^ I figured it out lol

  28. Outkast3r09
    • 2 years ago
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    chain rule?

  29. Outkast3r09
    • 2 years ago
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    i've never seen this lol.... suprisingly

  30. Outkast3r09
    • 2 years ago
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    i would've just chain ruled it

  31. monroe17
    • 2 years ago
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    it's x-2^{x}

  32. monroe17
    • 2 years ago
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    =1-2^{x}ln(2)

  33. abb0t
    • 2 years ago
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    wat? Lol

  34. monroe17
    • 2 years ago
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    nothing abb0t lol :)

  35. Outkast3r09
    • 2 years ago
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    is that an integration rule?

  36. Outkast3r09
    • 2 years ago
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    o gahd not the ln way of finding derivatives -.-

  37. monroe17
    • 2 years ago
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    lmao, ya --> ;/

  38. abb0t
    • 2 years ago
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    wait for ODE for differentiation w/ integration.

  39. Outkast3r09
    • 2 years ago
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    what do you mean? i've taken Diff Eq

  40. Outkast3r09
    • 2 years ago
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    suprisingly i've gone all the way through math not seeing anything like that lol

  41. Outkast3r09
    • 2 years ago
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    or i did and it was a longg time ago

  42. abb0t
    • 2 years ago
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    for exact equations. also in PDE. chain rule.

  43. Outkast3r09
    • 2 years ago
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    exact dquations are simple, \[Ndy=Mdx\] or i might have the backwards i always seem to do it

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