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Math Help - Simple Differentiation

  1. #1
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    Simple Differentiation

    Hello I have the question 'Find the equations of the tangents to these curves at correspnding points '

    I'm getting very frustrated becuase I cannot where I'm going wrong . Heres how i've been doing it.

    y=2/x , x=2 hence points (2,1/2)

    y=2x- (Is this where I went wrong?)

    y=-4x^-3

    dy/dx = 4*1/8
    =1/2

    y-1/2=4/8(x-2)

    y=1/2x-1/2


    Yet this isnt the line equation
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  2. #2
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    Hi there,

    You know the equation of a line given a gradient and point is y-y_{1}=m(x-x_{1})

    you also know:

    y=c \Rightarrow \frac{dy}{dx}=0

    y=kx \Rightarrow \frac{dy}{dx}=k

    y=kx^n \Rightarrow \frac{dy}{dx}=nkx^{n-1}

    Find m by finding the derivitive of y = \frac{2}{x^2} = 2x^{-2} when x is 2

    you have already found the point (2, \frac{1}{2})

    looking at your working it looks correct to me, except for the sign on m
    Last edited by sammy28; November 25th 2009 at 02:59 PM. Reason: added possible explanation
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  3. #3
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    Thats what I did though, and I doing the geometry right?
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  4. #4
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    check the sign on your gradient you should then get the correct line.
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  5. #5
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    Should this come under Pre-Calculus or Calculus?
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  6. #6
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    Quote Originally Posted by alibond07 View Post
    Hello I have the question 'Find the equations of the tangents to these curves at correspnding points '

    I'm getting very frustrated becuase I cannot where I'm going wrong . Heres how i've been doing it.

    y=2/x , x=2 hence points (2,1/2)

    y=2x- (Is this where I went wrong?)

    y=-4x^-3

    dy/dx = 4*1/8
    =1/2

    y-1/2=4/8(x-2)

    y=1/2x-1/2


    Yet this isnt the line equation
    Code:
    y=2x-   (Is this where I went wrong?)
     
    y=-4x^-3
     
    dy/dx = 4*1/8
            =1/2
    should be y=2x-

    dy/dx = -4x^-3

    dy/dx = - 4*1/8
    =- 1/2

    \rightarrowy-1/2=-4/8(x-2)


    \rightarrowy=-1/2x+3/2
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