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Finding the derivative of trig function!

  1. Oct 8, 2011 #1
    Find dy/dx of these functions:

    1. y=sin-1(1/3x)
    dy/dx=1/(√1-(1/3x)2 (1/3)
    dy/dx=1/(3√1-(1/9x2)

    However the book answer is 1/√9-x2). How did they get this?

    2. y=sin-1(1/x)
    dy/dx=1/√1-(1/x)2 (-1/x2)
    If i tried to simplified it by multiply it by x^2/x^2then
    dy/dx= -x^2/x4√x2-1 which equal -1/x2√x2-1 However the book answer is -1/absx√x2-1. Can you explain to me how they get this?

    3. find dy/dx by implicit differentiation
    sin-1(xy)=cos-1(x-y)
    so d/dx[sin^-1(xy)=cos^-1(x-y)]
    1/√1-(xy)2 (y+x) (dy/dx)=-1/√1-(x-y)2 (0)
    (y+x)/√1-(xy)^2 (dy/dx)=0
    dy/dx= (- y+x)/(√1-(xy)2)

    The formula i used for all these problems are
    d/dx[sin-1u]=1/√1-u2 (du/dx)
    d/dx[cos-1u]=-1/√1-u2 (du/dx)

    ~Thank you so much in advance. I really need help before my big test!
     
  2. jcsd
  3. Oct 8, 2011 #2
    For the first one, they just stuck the 3 in the denominator that came from the chain rule back into the radical.

    [tex]\frac{1}{3\sqrt{1-\frac{1}{9}x^{2}}}[/tex]
    [tex]\frac{1}{\sqrt{9(1-\frac{1}{9}x^{2})}}[/tex]
    [tex]\frac{1}{\sqrt{(9-\frac{9}{9}x^{2})}}[/tex]
    [tex]\frac{1}{\sqrt{(9-x^{2})}}[/tex]
     
    Last edited: Oct 8, 2011
  4. Oct 9, 2011 #3
    thank you what about the other two?
     
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