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monroe17

can someone help me step by step? Find the derivative 3x^(2)+e^(2y)=4y+tan(x)

  • one year ago
  • one year ago

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  1. Operative
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    solve for dx right?

    • one year ago
  2. monroe17
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    dy/dx

    • one year ago
  3. calculusfunctions
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    Have you learned implicit differentiation?

    • one year ago
  4. monroe17
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    I'm not good at it.

    • one year ago
  5. calculusfunctions
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    It's easy! Let me explain it to you in simple terms without getting wordy. If x and y are on the same side of the equal sign, then we say the function is written implicitly, and call it an implicit function. To find the derivative of an implicit function, take the derivative as you normally would but every time you differentiate y, multiply by dy/dx. Then solve for dy/dx. Example: x² + y³ = 5 2x + 3y²(dy/dx) = 0 3y²(dy/dx) = -2x dy/dx = -2x/3y² Understand @monroe17 ?

    • one year ago
  6. monroe17
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    I got it :) and when you have both x and y on each side you move the y terms to the left? and x on the right then divide?

    • one year ago
  7. calculusfunctions
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    Yes you isolate dy/dx and then solve for dy/dx. Good, you got it!

    • one year ago
  8. monroe17
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    for instance with my problem i would subtract 6x from both sides and it ends up on the right side of the equal sign. Then 4 is substracted to the right side.. getting sec^2(x)-6x/2e^(2y)-4?

    • one year ago
  9. calculusfunctions
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    Excellent!!

    • one year ago
  10. monroe17
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    yay, gosh finally.. imma try it for my next problem and see if i get it right.

    • one year ago
  11. calculusfunctions
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    Sure! I'll be on line for another hour or so, hence if you need anything, let me know.

    • one year ago
  12. monroe17
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    thanks!

    • one year ago
  13. calculusfunctions
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    Welcome! Good Luck!

    • one year ago
  14. monroe17
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    opps. I came to a problem.. ;/ my next question is 5y^2=3x^(2)y+e^(x) i did.. 10y(dy/dx)=6x(dy/dx)+e^(x) dy/dx(10y*y)=6x+e^(x) dy/dx=6x+e^(x)/10y*y and that's wrong.. what was my mistake??

    • one year ago
  15. zepdrix
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    Woops! When you took the derivative of the right side, it looks like you didn't apply the product rule on the 3x^2y term! :O

    • one year ago
  16. monroe17
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    ohhhh okay let me try that again.

    • one year ago
  17. calculusfunctions
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    Exactly as @zepdrix said!

    • one year ago
  18. calculusfunctions
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    @monroe17 remember you only multiply by dy/dx when you take the derivative of y, not every time you write y.

    • one year ago
  19. monroe17
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    oh i don't ?

    • one year ago
  20. monroe17
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    I got.. 10y(dy/dx)=3x^2+6x*y+e^(x)

    • one year ago
  21. calculusfunctions
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    Example: x² + 2xy - 4 = 6y 2x + 2y + 2x(dy/dx) = 6(dy/dx) Do you see what I did there?

    • one year ago
  22. monroe17
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    how do i know which term goes to which side for this problem.

    • one year ago
  23. Eda2012
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    \[5y ^{2} = 3x ^{2}y + e ^{x}\] \[10 y \frac{ dy }{ dx} = 3x ^{2 }\frac{ dy }{ dx } + 6xy + e ^{x}\] \[\frac{ dy }{ dx } = \frac{ 6xy + e ^{x} }{ 10y - 3x }\]

    • one year ago
  24. calculusfunctions
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    You know the product rule for derivatives, right?

    • one year ago
  25. calculusfunctions
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    @Eda2012 I am trying to teach her. I do not appreciate you just posting the solution. How is she supposed to learn to solve problems independently if you just give her the answers?

    • one year ago
  26. monroe17
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    ^thank you! lol

    • one year ago
  27. monroe17
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    the product rule is f(x)g'(x)+g(x)f'(x) right?

    • one year ago
  28. calculusfunctions
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    Correct! Thus if you have a term of 3x²y for example, then your f = 3x² and g = y. And you multiply by dy/dx iff you differentiate y. Understand?

    • one year ago
  29. calculusfunctions
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    If f = 3x² and g = y, then f ' = 6x and g' = dy/dx Do you see?

    • one year ago
  30. monroe17
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    so.. 10y(dy/dx)=3x^2(dy/dx)+y(6x)+e^(x) 10y(dy/dx)=3x^(2)+6xy+e^(x)

    • one year ago
  31. monroe17
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    why does 3x^2 go to the left side? I though you make all x to one side and y's the other?

    • one year ago
  32. calculusfunctions
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    (fg)' = fg' + gf' (3x²y)' = 3x²(dy/dx) + 6xy Do you see your mistake when you differentiated 3x²y on the right side?

    • one year ago
  33. Eda2012
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    @calculusfunctions sometimes you have to show some works and then explain it...sorry...

    • one year ago
  34. monroe17
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    oh yeah my bad. I mixed them up.

    • one year ago
  35. calculusfunctions
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    @Eda2012 so you show examples and explain the lessons not just give out the answers. I am a teacher so you really don't need to tell me how to do my job!

    • one year ago
  36. Eda2012
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    @calculusfunctions ok...next time i won't do it again...sorry

    • one year ago
  37. calculusfunctions
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    @monroe17 do you understand now?

    • one year ago
  38. monroe17
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    ^ calculus it's okay, I ignored it :) why does 3x^2 go to the left side? I though you make all x to one side and y's the other?

    • one year ago
  39. calculusfunctions
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    Because it is 3x²(dy/dx) and you're isolating dy/dx.

    • one year ago
  40. monroe17
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    ... im confused

    • one year ago
  41. calculusfunctions
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    OK Do you want me to show you step by step?

    • one year ago
  42. monroe17
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    let me look at it real quick.

    • one year ago
  43. calculusfunctions
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    Since @eda2012 already posted the correct solution, go ahead and observe it and then tell me which of her steps confuses you, if any. Alright?

    • one year ago
  44. monroe17
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    okay, give me a little time

    • one year ago
  45. calculusfunctions
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    Sure! Someone else is asking for my help so I'll go and get them started. You call me when you're ready. Take your time.

    • one year ago
  46. monroe17
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    I'm sorry that took forever, I had a lab report to complete. But, I re-looked over the problem and it seems to make a little more sense. Thank you so much for your help!

    • one year ago
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