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tmagloire1

  • one year ago

Can anyone help me on this problem regarding implicit differentiation? Thanks http://prntscr.com/7u700y

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  1. DanJS
    • one year ago
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    when differentiating y, remember to use the chain rule...

  2. DanJS
    • one year ago
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    For example, \[\frac{ d }{ dx }[y^3] = \frac{ d }{ dy }[y^3]*\frac{ dy }{ dx }\]

  3. DanJS
    • one year ago
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    Basic chain rule for this \[\frac{ d }{ dx } = \frac{ d }{ dy }*\frac{ dy }{ dx }\]

  4. DanJS
    • one year ago
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    Following the above example, \[\frac{ d }{ dx }[y^3] = 3y^2*\frac{ dy }{ dx }\]

  5. DanJS
    • one year ago
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    Do that for the y terms, and differentiate the x terms normally, then solve for dy/dx

  6. tmagloire1
    • one year ago
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    Okay I'll try it and tell you what I get.

  7. DanJS
    • one year ago
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    k

  8. tmagloire1
    • one year ago
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    Okay I am actually still very confused. Am i supposed to find the second derivative of 5x^2 + y4 = −9 and then use implicit differentiation?

  9. DanJS
    • one year ago
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    Yes you want to find the second derivative of the given thing with respect to X, but to take the derivative w.r.t. X of the Y term, you have to use that chain rule... here ill type it out

  10. DanJS
    • one year ago
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    \[\frac{ d }{ dx }[5x^2]+\frac{ d }{ dx }y^4 = \frac{ d }{ dx }[-9]\] \[\frac{ d }{ dx }[5x^2]+\frac{ d }{ dy }y^4*\frac{ dy }{ dx } = \frac{ d }{ dx }[9]\]

  11. DanJS
    • one year ago
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    now take derivative of y term with respect to y like normal, but you have to put the y' on there too from the chain rule.

  12. DanJS
    • one year ago
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    \[\frac{ d }{ dx }[5x^2]+\frac{ d }{ dy }y^4*\frac{ dy }{ dx } = \frac{ d }{ dx }[-9]\] \[10x^2+4y^3*\frac{ dy }{ dx } = 0\]

  13. DanJS
    • one year ago
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    Solve that for dy/dx and that is your first derivative...

  14. DanJS
    • one year ago
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    Or just remember, if you are taking the derivative of a Y term, tag on a dy/dx next to it ...basically

  15. tmagloire1
    • one year ago
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    Okay for the first derivative I got : dy/dx= -10x^2 / 4y^3

  16. tmagloire1
    • one year ago
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    And then take the derivative again?

  17. DanJS
    • one year ago
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    So you get, \[\frac{ dy }{ dx } = \frac{ -10x^2 }{ 4y^3 }\] right

  18. DanJS
    • one year ago
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    This time you need the product or the quotient rule and the chain rule...

  19. DanJS
    • one year ago
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    I can start typing it out...

  20. tmagloire1
    • one year ago
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    Im using the quotient rule

  21. DanJS
    • one year ago
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    \[\frac{ d }{ dx }\frac{ dy }{ dx } = \frac{ d^2y }{ dx^2 }\] k, ill start typing it

  22. tmagloire1
    • one year ago
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    I got : -80xy^3 +120x^2y^2 / (4y^3)^2

  23. DanJS
    • one year ago
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    \[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx}\frac{ -5x^2 }{ 2y^3 }\]

  24. tmagloire1
    • one year ago
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    Oh I think I took the derivative of the wrong thing?

  25. tmagloire1
    • one year ago
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    So it would be -20 when you plug in x and y

  26. DanJS
    • one year ago
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    \[\frac{ d^2y }{ dx^2 }=\frac{ 2y^2*-10x - [-5x*6y^2*\frac{ dy }{ dx } ]}{ 4y^6 }\]

  27. DanJS
    • one year ago
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    That is the quotient rule , using the chain rule on the y term in the numorator.

  28. tmagloire1
    • one year ago
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    So am I supposed to plug in x and y into which equation?

  29. DanJS
    • one year ago
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    yes, but notice the second derivative has a term of the first derivative in the numerator... but you know what dy/dx is , so sub that in first

  30. DanJS
    • one year ago
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    \[\frac{ dy }{ dx } = \frac{ -5x^2 }{ 2y^3 }\] reduced fraction... Sub that in for dyy/dx in the second derivative then plug in the x and y and solve...

  31. DanJS
    • one year ago
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    Less writing overall if you want to use y' and y'' instead of dy/dx things

  32. DanJS
    • one year ago
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    goodluck, i dont want to calculate it.

  33. Jhannybean
    • one year ago
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    \[\large \color{blue}{\checkmark}\]

  34. tmagloire1
    • one year ago
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    I got -310

  35. tmagloire1
    • one year ago
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    Thanks for all the help!

  36. DanJS
    • one year ago
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    yw, just get that Leibanitz notation dy/dx crap down, it makes the chain rule and this stuff easier i think .. goodluck

  37. tmagloire1
    • one year ago
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    Okay I will try using it more.

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