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Find dy/dx by implicit differentiation

  1. Oct 14, 2007 #1
    Question:
    Find dy/dx by implicit differentiation
    4x^2 + 3xy - y^2 = 6

    Attempt:
    Ok, just a forewarning that I suck at differentiations, limits, what-not, so..
    Following my textbook, it says I should Differentiate both sides of the equation
    So...
    (d/dx)(4x^2 + 3xy - y^2) = (d/dx)(6)
    (d/dx)(6) = 0, so I have to solve the derivatives of the other side..
    Now, this is the part that I'm unsure about, so please guide me here:
    (d/dx)(4x^2) = 8x right? Since I bring down the two and subtract 1
    Ok, so.. (d/dx)(3xy), using the chain rule which is f(g(x)) = f'(g(x))*g'(x), I figured:
    y(...) is the outer function and 3x is the inner funtion
    thus, making it (y(3x))' * (3x)' = (dy/dx)(3x) * 1???

    Can someone please clarify that last part I wrote for me? Also, it'd be great if someone can clarify the difference and meaning between dy/dx, d/dx, and d/dy..
    Then I'll try and move on XD
    Please and thank you for your time XD

    PS: I do realise I haven't got through the whole question, but I will, once the above is clarified as correct, etc. :D
     
  2. jcsd
  3. Oct 14, 2007 #2
    you would use the product rule for 3xy, the chain rule applies to y since y is a function of x.
    so you'd have f(x)g(x) = f(x)g'(x)+g(x)f'(x) where either f or g is y. and the derivative of y will be dy/dx.

    dy/dx is the derivative of y with respect to x. d/dx and d/dy are operators I think and you use them on a function like y to take the derivative of that function with respect to x or y.
     
  4. Oct 14, 2007 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that is correct.

    No, there is no "inner"or "outer" function- this is not composition of functions, it is just a product of variables- use the product rule: (3xy)'= (3y)(x)'+ (3x)y'= 3y+ 3xy' since you are differentiating with respect to x.

    Okay, I think I did "clarify" above. As for the second question, dy/dx means the derivative of y with respect to x- presumably, y is a function of x- if not that derivative is just 0. d/dx means the derivative of some function with respect to x. Usually that is used when the function is just to long to fit reasonably inside the derivative
    [tex]d/dx (x^3- 3sin(x)+ xe^x- 3xcos(x)+ x^3- 3x^2+ 4x)[/itex]
    rather than
    [tex]\frac{d (x^3- 3sin(x)+ xe^x- 3xcos(x)+ x^3- 3x^2+ 4x)}{dx}[/tex]

    d/dy is the same thing except that you are differentiating with respect to the variable y. Presumably now the function being differentiated is a function of y. If there are any other variables in the function, they should be also functions of y. You probably haven't had "partial derivatives" yet. That's what you would use if you had a function of two or more independent variable.
     
  5. Oct 14, 2007 #4
    Oh ok
    So for the part 3xy , does the product rule have to be applied twice then?
    As in
    (3)(x)'+(x)(3)' and then (3x)(y)' + (y)(3x)'?

    Edit: No wait, scratch that
    It's just (3x)(y)' + (y)(3x)' then, right?
    I'm not exactly sure how you did that
     
    Last edited: Oct 14, 2007
  6. Oct 14, 2007 #5
    yes but you can simplify (3x)'.

    for the (3xy)' HallsofIvy just used the product rule, you used it too your answers are exactly the same when you simplify (3x)'.
     
  7. Oct 14, 2007 #6
    [tex]3[xy][/tex]

    [tex]3[xy'+y][/tex]

    [tex]3xy'+3y[/tex]
     
  8. Oct 14, 2007 #7
    Ok, gotcha :D
    I have another quick question, would y' become dy/dx then? That's what dy/dx means, right?
     
  9. Oct 14, 2007 #8
    yes y'=dy/dx
     
  10. Oct 14, 2007 #9
    :D Awesome
    So in order to find (d/dx)(y^2), now I have to use the chain rule, right?
    (...)^2 would be outer and y would be inner, so..

    f'(g(x))*g'(x) = .... 2(y)*y' then?
    Since the derivative of ^2, I bring down the two, but I don't do anything to the y inside, right?
     
  11. Oct 14, 2007 #10
    right (y^2)' = 2yy'

    and yes you can't do anything to y' since y' is a function of x that you don't know since it's defined implicitly.
     
  12. Oct 14, 2007 #11
    :D
    Ok, so now that I've got the d/dx of the values, I assume I'll have to isolate dy/dx
    So..

    8x + 3x(dy/dx) + 3y - 2y(dy/dx) = 0
    then, putting dy/dx on one side..
    2y(dy/dx) - 3x(dy/dx) = 8x + 3y
    meaning
    (dy/dx)(2y - 3x) = (8x + 3y)
    then
    (dy/dx) = (8x + 3y)/(2y - 3x)
    :D
    Do you think I am allowed to leave it like that?
     
  13. Oct 14, 2007 #12
    yep that's right, since you can't reduce the fraction anymore.
     
  14. Oct 14, 2007 #13
    YAY :D
    Thank you so much bob for sticking with me so long :D :D:D:D:D:D
     
  15. Oct 14, 2007 #14
    no problem, glad I could help
     
  16. Oct 14, 2007 #15
    lols, omg.
    I'm on my next question, and I'm sort of stuck.. again :(

    So here's the next question:
    Find dy/dx by implicit differentiation.
    x^2y^2 + x sin(y)= 9


    Ok, so I know I'm to take d/dx on both sides
    But if it's (d/dx) of (x^2y^2), how would I do it?
    Would I use the chain rule for x^2 and y^2 separately, and then use the product rule?
     
  17. Oct 14, 2007 #16
    no you would use the product rule first, and when you get to (y^2)' you use the chain rule.
     
  18. Oct 14, 2007 #17
    Oh ok, so assuming I did it correctly, it'd be
    x^2(y^2)' + y^2(2x)
    and (y^2)' = 2y(y)'

    I'm unsure about the d/dx of x sin(y).. would I be using the chain rule and then the product rule?
     
  19. Oct 14, 2007 #18
    product then chain rule for (sin y)' since you have xsin(y) you need to be doing the derivative of just sin(y) before you can use the chain rule.
     
  20. Oct 14, 2007 #19
    Right!
    Ok, so I solved
    x(sin y)' + (sin y)(x)'
    Which becomes x(cos y*y') + (sin y)
    Since (sin y)' = (cos y)(y') and x' = 1
    And now, isolating dy/dx...

    2yx^2(dy/dx) + x(cos y)(dy/dx) + (sin y) = 0
    then 2yx^2(dy/dx) + x(cos y)(dy/dx) = -(sin y)
    (dy/dx)(2yx^2 + x(cos y)) = -(sin y)

    So (dy/dx) = -(sin y)/(2yx^2 + x(cos y)) right?
    I inputted it in my online math homework site place, and it's wrong ToT
    What did I do wrong?
     
  21. Oct 14, 2007 #20
    when you took the derivative of x^2y^2 you only have down the 2yy'x^2 part, you're missing the +2xy^2.
     
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