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monroe17 Group Title

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

  • 2 years ago
  • 2 years ago

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  1. Operative Group Title
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    solve for dx right?

    • 2 years ago
  2. monroe17 Group Title
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    dy/dx

    • 2 years ago
  3. calculusfunctions Group Title
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    Have you learned implicit differentiation?

    • 2 years ago
  4. monroe17 Group Title
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    I'm not good at it.

    • 2 years ago
  5. calculusfunctions Group Title
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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 ?

    • 2 years ago
  6. monroe17 Group Title
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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?

    • 2 years ago
  7. calculusfunctions Group Title
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    Yes you isolate dy/dx and then solve for dy/dx. Good, you got it!

    • 2 years ago
  8. monroe17 Group Title
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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?

    • 2 years ago
  9. calculusfunctions Group Title
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    Excellent!!

    • 2 years ago
  10. monroe17 Group Title
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    yay, gosh finally.. imma try it for my next problem and see if i get it right.

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

    • 2 years ago
  12. monroe17 Group Title
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    thanks!

    • 2 years ago
  13. calculusfunctions Group Title
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    Welcome! Good Luck!

    • 2 years ago
  14. monroe17 Group Title
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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??

    • 2 years ago
  15. zepdrix Group Title
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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

    • 2 years ago
  16. monroe17 Group Title
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    ohhhh okay let me try that again.

    • 2 years ago
  17. calculusfunctions Group Title
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    Exactly as @zepdrix said!

    • 2 years ago
  18. calculusfunctions Group Title
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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.

    • 2 years ago
  19. monroe17 Group Title
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    oh i don't ?

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

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

    • 2 years ago
  22. monroe17 Group Title
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    how do i know which term goes to which side for this problem.

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

    • 2 years ago
  24. calculusfunctions Group Title
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    You know the product rule for derivatives, right?

    • 2 years ago
  25. calculusfunctions Group Title
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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?

    • 2 years ago
  26. monroe17 Group Title
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    ^thank you! lol

    • 2 years ago
  27. monroe17 Group Title
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    the product rule is f(x)g'(x)+g(x)f'(x) right?

    • 2 years ago
  28. calculusfunctions Group Title
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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?

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

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

    • 2 years ago
  31. monroe17 Group Title
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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?

    • 2 years ago
  32. calculusfunctions Group Title
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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?

    • 2 years ago
  33. Eda2012 Group Title
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    @calculusfunctions sometimes you have to show some works and then explain it...sorry...

    • 2 years ago
  34. monroe17 Group Title
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    oh yeah my bad. I mixed them up.

    • 2 years ago
  35. calculusfunctions Group Title
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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!

    • 2 years ago
  36. Eda2012 Group Title
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    @calculusfunctions ok...next time i won't do it again...sorry

    • 2 years ago
  37. calculusfunctions Group Title
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    @monroe17 do you understand now?

    • 2 years ago
  38. monroe17 Group Title
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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?

    • 2 years ago
  39. calculusfunctions Group Title
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    Because it is 3x²(dy/dx) and you're isolating dy/dx.

    • 2 years ago
  40. monroe17 Group Title
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    ... im confused

    • 2 years ago
  41. calculusfunctions Group Title
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    OK Do you want me to show you step by step?

    • 2 years ago
  42. monroe17 Group Title
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    let me look at it real quick.

    • 2 years ago
  43. calculusfunctions Group Title
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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?

    • 2 years ago
  44. monroe17 Group Title
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    okay, give me a little time

    • 2 years ago
  45. calculusfunctions Group Title
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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.

    • 2 years ago
  46. monroe17 Group Title
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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!

    • 2 years ago
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