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monroe17

  • 2 years ago

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

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  1. Operative
    • 2 years ago
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    solve for dx right?

  2. monroe17
    • 2 years ago
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    dy/dx

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

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

  5. calculusfunctions
    • 2 years ago
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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 ?

  6. monroe17
    • 2 years ago
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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?

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

  8. monroe17
    • 2 years ago
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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?

  9. calculusfunctions
    • 2 years ago
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    Excellent!!

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

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

  12. monroe17
    • 2 years ago
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    thanks!

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

  14. monroe17
    • 2 years ago
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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??

  15. zepdrix
    • 2 years ago
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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

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

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

  18. calculusfunctions
    • 2 years ago
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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.

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

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

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

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

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

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

  25. calculusfunctions
    • 2 years ago
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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?

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

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

  28. calculusfunctions
    • 2 years ago
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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?

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

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

  31. monroe17
    • 2 years ago
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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?

  32. calculusfunctions
    • 2 years ago
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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?

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

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

  35. calculusfunctions
    • 2 years ago
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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!

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

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

  38. monroe17
    • 2 years ago
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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?

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

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

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

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

  43. calculusfunctions
    • 2 years ago
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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?

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

  45. calculusfunctions
    • 2 years ago
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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.

  46. monroe17
    • 2 years ago
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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!

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