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Quick Tangent Line Approximation (Derivatives)

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data
    The tangent line through given points (pi/4,pi/4) m=1

    y= cos(y)cos(x)/sin(x)sin(y)


    3. The attempt at a solution
    dy/dx= d/dx[cos(y)cos(x)/sin(x)sin(y)]

    First use quotient rule ?
    vu'-uv'/v^2

    v= sin(x)sin(y)
    v'= do i need to use product rule? for product rule I got sin(x)*cos(x)+sin(x)*cos(x)
    but if I do it normal way, I get cos(x)*sin(y)
    u= cos(y)cos(x)
    u'= -sin(x)cos(y)

    Quotient Rule:
    sin(x)sin(y)*-sin(x)cos(y) - cos(y)cos(x)*cos(x)*sin(y)
    sin(x)sin(y)^2


    I just need to know if I'm going in the right direction. I'm pretty sure I know what to do for the point slope formula since its basic algebra. But it's this beginning part that I have problems with ! Thanks in advance :)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 9, 2008 #2

    Dick

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    Yes, you need to use the product rule. You also need to use the chain rule. d/dx(cos(y))=-sin(y)*dy/dx. You might find it easier to rearrange your original function into y*tan(y)=tan(x). Now take d/dx of both sides. You'll get more than one dy/dx. Treat it as a unknown and solve for it. This is called implicit differentiation.
     
  4. Nov 9, 2008 #3
    Thanks ! I never thought about changing it ...

    I followed your suggestion and took d/dx of both sides

    y*tan(y) use product rule-->
    u=y u'= dy/dx
    v=tan(y) v'= sec^2(y) * dy/dx ?

    y*sec^2(x)*dy/dx+ tan(y)*dy/dx = sec^2(x)

    Is this correct ?
     
  5. Nov 9, 2008 #4

    Dick

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    That's what I got.
     
  6. Nov 9, 2008 #5

    Dick

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    Oops. You accidentally wrote sec^2(x) on the left side instead of sec^2(y), right?
     
  7. Nov 9, 2008 #6
    oops yes, sorry . too used to using x. yes i meant sec^2(y) !

    how do i go about eliminating one of the dy/dx ? What are the next steps?
     
  8. Nov 9, 2008 #7

    Dick

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    Put x=y=pi/4 and solve for dy/dx. Or just solve for dy/dx then put x=y=pi/4. Your choice.
     
  9. Nov 9, 2008 #8
    Never mind. I found the answer. thanks again for the help :)
     
    Last edited: Nov 9, 2008
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