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Homework Help: Find the equation of the tangent line of the curve

  1. Mar 7, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the equation of the tangent line to the curve ##\ xy^2 + \frac 2 y = 4## at the point (2,1).
    Answer says ##\ y-1 = -\frac 1 2(x-2)##
    And with implicit differentiation I should have gotten ##\frac {dy} {dx}= -\frac {y^2} {2xy-\frac {2} {y^2}}##
    2. Relevant equations
    ##\ y-y_1 = m(x-x_1)##

    3. The attempt at a solution
    ##\ y-1 = m(x-2)##

    ##\frac d {dx}(xy^2+\frac 2 y) =\frac d {dx}(4)##

    ##\frac d {dx}(xy^2)+\frac d {dx}(\frac 2 y) = 0##

    ##2xy+x⋅\frac {dy} {dx}-\frac {2} {y^2}⋅ \frac {dy} {dx} = 0##

    ##x⋅\frac {dy} {dx}-\frac {2} {y^2}⋅ \frac {dy} {dx} = 2xy##

    ##\frac {dy} {dx}(x-\frac {2} {y^2})= 2xy##

    ##\frac {dy} {dx}= -\frac {2xy} {x-\frac {2} {y^2}}##
    Subbing in my x and y values gives me a slope of 4.
    If anyone can help show me what I did wrong, I would really appreciate it. Also this is the first time using LaTeX Primer so bare with me.
     
  2. jcsd
  3. Mar 7, 2017 #2

    FactChecker

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    Check this. It looks wrong to me.
     
  4. Mar 7, 2017 #3
    ##2xy+x⋅\frac {dy} {dx}-\frac {2} {y^2}⋅ \frac {dy} {dx} = 0## is this supposed to be
    ##x +2xy⋅\frac {dy} {dx}-\frac {2} {y^2}⋅ \frac {dy} {dx} = 0##?
     
  5. Mar 7, 2017 #4

    FactChecker

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    Methodically applying the product rule to d/dx (x⋅y2) gives
    d/dx (x⋅y2) = (d/dx x)⋅y2 + x⋅(d/dx y2) = 1⋅y2 + x⋅2⋅y⋅dy/dx = y2 + 2xy⋅dy/dx
    I don't see this in your calculations.
     
  6. Mar 7, 2017 #5
    So ##\frac d {dx}(xy^2)+\frac d {dx}(\frac 2 y) = 0## should turn into
    ##x\frac {d} {dx} (y^2)+ x⋅\frac d {dx} y^2- \frac 2 {y^2}(\frac {dy} {dx}) = 0## because I don't see the derivative of ##\frac 2 y##?
     
    Last edited: Mar 7, 2017
  7. Mar 8, 2017 #6

    Mark44

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    This part -- ##x\frac {d} {dx} (y^2)+ x⋅\frac d {dx} y^2## -- is wrong. You are not applying the product rule correctly. You have written the same thing twice.
     
  8. Mar 8, 2017 #7
    Ohhhh, I should have done this?
    ##\frac {d} {dx} (1⋅y^2)+\frac {dy} {dx}2y⋅x -\frac {dy} {dx} \frac 2 y^2= 0##
    then I would get ##-\frac {y^2} {2xy-\frac 2 {y^2}}## ?
     
  9. Mar 8, 2017 #8

    Mark44

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    No, that's not it either. Here's how the product rule works.
    ##\frac {d} {dx} (f(x) g(x)) = \frac {d} {dx}(f(x)) \cdot g(x) + f(x) \cdot \frac {d} {dx}(g(x)) = f'(x)g(x) + f(x)g'(x)##
    Let's start from the beginning.
    ##\frac {d} {dx} (xy^2) = (\frac {d} {dx} x) \cdot y^2 + x \frac {d} {dx}(y^2)##
    Can you continue? You'll need to use the chain rule to evaluate the derivative on the right.
     
  10. Mar 8, 2017 #9

    Ray Vickson

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    No, that is even worse.

    Go back to the beginning, and use the product rule carefully, doing it one step at a time. Check your work as you go along.
     
  11. Mar 8, 2017 #10
    ##\frac {d} {dx} x⋅y^2+\frac {dy} {dx}2y⋅x -\frac {dy} {dx} \frac 2 {y^2}= 0##
    then I would get ##-\frac {y^2} {2xy-\frac 2 {y^2}}##
    I thought that I did the product rule.
    f(x) = x
    f'(x) = 1
    g(x) = y2
    g'(x) = 2y
    I just did the product rule as f'g⋅g'f.
    After the product rule I add the ##\frac 2 y ## but the derivative of this turns it into ##-\frac 2 {y^2}##? I can't seem to see what I've done wrong.
     
  12. Mar 8, 2017 #11

    Ray Vickson

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    It does not turn into ##-2/y^2##, it turns into ##-2 y'/y^2##.
     
  13. Mar 8, 2017 #12

    FactChecker

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    No. Don't forget that a change of x forces a change of y because of the initial equation. The chain rule gives g'(x) = 2y ##\frac {dy} {dx}##
    Same problem here. Use the chain rule and include ##\frac {dy} {dx}##.
     
  14. Mar 8, 2017 #13
    I think I'll just have to watch more videos on implicit differentiation because I'm lost.
     
  15. Mar 8, 2017 #14
    Ok, let me see if I've done this properly. The product rule - ##(x\frac{dy}{dx}⋅2xy)+(1⋅y^2)##. Now for the ##\frac 2 y ## do I use the quotient rule?
     
  16. Mar 8, 2017 #15

    Ray Vickson

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    You could use the quotient rule, but it is a whole lot easier just to use the chain rule.
     
  17. Mar 8, 2017 #16
    Ok so the chain rule is ##f'((g(x))⋅g'(x)##, correct? I'm a little confused as to what I am subbing in where.
     
  18. Mar 8, 2017 #17

    Mark44

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    Here's the chain rule for the problem you're having, using Leibniz notation.
    ##\frac d{dx}(y^2) = \frac d {dy}(y^2) \cdot \frac{dy}{dx}##
    Can you continue?
     
  19. Mar 8, 2017 #18

    FactChecker

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    You can use the derivative of a power and the chain rule to calculate 2/y.

    2/y is really 2⋅y-1 and y is really a function of x, so it is 2⋅y(x)-1
    Therefore ##\frac {d} {dx}##(2/y)
    = 2⋅##\frac {d} {dx}##(y(x)-1)
    = 2⋅(-1)y(x)-2⋅y'(x)
    = (-2/y2)⋅##\frac {dy} {dx}##
     
  20. Mar 9, 2017 #19

    cristo

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    It helps if you try to understand why you are doing what you are doing.

    So, to differentiate [itex]\frac{2}{y}[/itex] with respect to x, you want to first differentiate with respect to y: [itex]\frac{d}{dy}\Big(\frac{2}{y}\Big)[/itex] and then multiply by dy/dx. So you have [itex]\frac{d}{dx}\Big(\frac{2}{y}\Big)=\frac{d}{dy}\Big(\frac{2}{y}\Big)\cdot\frac{dy}{dx}[/itex]
     
  21. Mar 9, 2017 #20
    Ohhh ok so it should look something like this?

    ##\frac {d}{dx}(xy^2+\frac 2 y )= \frac d {dx}(4)##
    ##\frac {d}{dx}(xy^2)+\frac d{dx}\frac 2 y=0##
    ##(x\frac d{dx}⋅2xy)+(1⋅y^2)+(\frac d{dx}\frac 2 y )= 0##
    ## 2xy⋅\frac {dy}{dx}+y^2-\frac 2 {y^2}⋅\frac {dy}{dx}=0##
    ## \frac {dy}{dx}(2xy- \frac 2 {y^2}) = -y^2##
    ## \frac {dy}{dx} = - \frac {y^2} {2xy- \frac 2 {y^2}}##
    Have I done this correctly?
     
  22. Mar 9, 2017 #21

    Mark44

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    No. The first two lines are correct, but things fall apart in the third line.
    In differentiating ##xy^2## you are have a mistake here:
    For this part of the product rule you should have ##x\frac d {dx}(y^2)##. You'll need to use the chain rule as I described in a previous post to finish this part.

    Also, do not write ##\frac d{dx}⋅2xy##. You are not multiplying ##\frac d {dx}## and 2xy, which is what the dot (⋅) signifies.
    In the line below the one I'm talking about, it looks like you corrected your mistake.
     
  23. Mar 9, 2017 #22

    FactChecker

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    I would say that you need to work on mastering the basic rules of derivatives some more (product, chain, quotient). Almost all problems are solved using a combination of those. Until you get them mastered, there is little chance of getting more complicated combinations correct.
     
  24. Mar 9, 2017 #23
    Yes, I've been trying to master the basics. I'm hoping that khan academy can help with that.
     
  25. Mar 9, 2017 #24

    Ray Vickson

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    Why not purchase an actual BOOK on the subject? You can learn a lot from (for example) the Schaum's Outline series. Also: there are lots of free calculus textbooks available on-line, as well as tutorials and course notes, etc.
     
  26. Mar 9, 2017 #25
    I did not know of any books available. The course I'm doing is not very good. Thanks for the info I'm more than willing to purchase a book if it'll help.
     
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