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Implicit Differentiation with Trig

  1. Oct 20, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the slope of the tangent line to x tan y = y - 1 when y = pi/4

    2. Relevant equations

    3. The attempt at a solution

    I can't seem to get the derivative. Here's what I do.

    First I used the product rule the left side of the equation and got sec^2 x dy/dx + tan y = dy/dx.

    When I simplify, I get dy/dy over dy/dx, which just reduces to 1 so I'm having trouble with getting dy/dx on it own.
  2. jcsd
  3. Oct 20, 2007 #2
    you miswrote it


    but either way you should have been able to algebraically solve for dy\dx

    what would your next step be to find x?
    Last edited: Oct 20, 2007
  4. Oct 20, 2007 #3
    I derived it and I got:

    dy/dx = -tany/(xsec^2 - 1).

    Is that the right derivative?
  5. Oct 20, 2007 #4
  6. Oct 20, 2007 #5
    Thanks for the help, people!

    I need help with the tangent line now though. I have to find it when y = pi/4.
    I'm not sure how to approach it because I don't have the value for x.
  7. Oct 20, 2007 #6
    you can easily find it, you're given your y value. you also have your original equation, correct? where does the tangent line pass through?
  8. Oct 20, 2007 #7
    It passes through pi/4.

    Here's my equation:

    dy/dx = -tan(pi/4)/(xsec^2 - 1)

    I have two unknowns so not sure how to find it.
  9. Oct 20, 2007 #8
    [tex]x\tan{y}=y-1[/tex] @ [tex](x,\frac{\pi}{4})[/tex]
    Last edited: Oct 20, 2007
  10. Oct 20, 2007 #9
    For x, I got:

    pi/4 - 1

    When I find the slope and plug these in, I don't get what to do with the x and sec^2. Is x part of the sec or not?
  11. Oct 20, 2007 #10
    that's not the X value i got, check it again


    the derivative is the slope of your tangent line at your (x,y) so once you find both of those, you can find [tex]\frac{dy}{dx}[/tex]
  12. Oct 20, 2007 #11
    Thanks for the reply. I solved for x and still got: pi/4 - 1.

    I'm not sure how to get the derivative by subbing in for dy/dx. What should I do with the sec^2. Is x part of the sec^2 function or is it a coefficient?
    Last edited: Oct 20, 2007
  13. Oct 20, 2007 #12
    lol i'm so sorry, had a dumb moment. [tex]\frac{\pi}{4}=1[/tex]

    i make too many mistakes!

    [tex]y'=\frac{dy}{dx}=\frac{\tan{y}}{1-x\sec^{2}y}[/tex] @ (x,y) = m = slope
    Last edited: Oct 20, 2007
  14. Oct 21, 2007 #13
    I got the same as numerator as you for the derivative but my denominator is xsec^2 y - 1
  15. Oct 21, 2007 #14
    it's equivalent
  16. Oct 21, 2007 #15
    This is the reply for bond girl007

    differentiate the two sides of the equation:
    d/dx(xtany) =d/dx(y-1) Lows of differentiation


    make dy/dx in one side: tany=(1-sec^2y)dy/dx

    using laws of trigonometrics: tany=(-tan^2y)dy/dx

    we get: 1=-tanydy/dx



    Hence: dy/dx=-1
  17. Oct 21, 2007 #16
    [tex][x\tan{y}=y-1] \neq [\frac{dy}{dx}=\sec^{2}y\frac{dy}{dx}+\tan{y}][/tex]

    [tex][x\tan{y}=y-1] \mbox{should equal, if i'm not mistaken b/c it's almost 2am} [x\sec^{2}y\frac{dy}{dx}+\tan{y}=\frac{dy}{dx}][/tex]

    check plz!
    Last edited: Oct 21, 2007
  18. Oct 21, 2007 #17
    Indeed, ali1982 forgot the x in his equation. The result x=pi/4-1 is indeed correct. Now it is not difficult to find the slope. It is allready derived as dy/dx=tan(y)/[1-x*sec^2(y)]. The numbers x and y have been obtained, just put them in here. I think yo should get something like 2/(6-pi) approx. 0.7
  19. Oct 21, 2007 #18
    thanYes, I forgot the x.When we differentiate, the term : sec^2ydy/dx must be multiplied by x ,and this term will be : xsec^2ydy/dx.
    Thanks for correction.
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