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Math Help - Differentiation problems

  1. #1
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    Differentiation problems

    Hey i am having a bit of trouble with these differentiation problems;

    1. Find the equation to the tangent to y=e^-x, at the point where x = 1
    2. Find the equation to the tangent to y= ln(2-x) at the point where x = -1
    3. The tangent at x = 1 to y = x^2e^x cuts the x and y axis at A and B respectively. Find the coordinate of A and B
    4. Find the equation of the normal to y = ln √ x at the point where y = -1

    Im not sure about any of these but for the first 2 i think you have to differentiate first but im nt really sure..

    thanks!
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  2. #2
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    Ive managed to do the first 2 questions

    Any help on the last 2 would be appreciated
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  3. #3
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    Solution

    Q.1 : y=e^-x equation of tangent at point of abcissa x = 1

    1. y'=-e^-x
    2. y'(1)=-e
    3. y(1)=e^-1
    then the equation of the tangent is : Yt=y'(1) ( X-1 ) + y(1)
    Yt=-e(x-1)+e^-1 ; Yt=-ex+(e+e^-1).

    Q.2 : y= ln(2-x) x=-1

    1. y'= ( -1 ) / ( 2-x )
    2. y'(-1) = -1/3
    3. y(-1) = ln(3)
    Yt=y'(-1) ( X+1 ) + y(-1)
    Yt= (-1/3) ( X+1 ) + ln(3)
    Yt= (-1/3)X+(ln(3) -1/3 )

    Q.3

    first you get the equation of the tangent just as Q1 and Q2 the general form of the equation of the tangent Yt to a point of absissa Xt on a graph of equation Y is : Yt = Y'(Xt) ( X-Xt ) + Y(Xt)

    1. the equation of the tangent here is : Yt=(3e)X-2e
    2. the point A is where Yt=0 , => A(2/3,0)
    3. the point B is where X=0 , ==> B(0,-2e)

    Q.4 i didnt undestand what do u mean by the normal to y=ets ...
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  4. #4
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    Quote Originally Posted by Jampop View Post
    [snip]
    4. Find the equation of the normal to y = ln √ x at the point where y = -1

    Im not sure about any of these but for the first 2 i think you have to differentiate first but im nt really sure..

    thanks!
    You need the coordinates of the point you're finding the normal at. You're given the y-coordinate. To find the x-coordinate, solve -1 = \ln \sqrt{x} for x.

    To get the gradient, recall that m_{normal} = - \frac{1}{m_{tangent}} so you first need to find \frac{dy}{dx} at the value of x found above.

    Now you have a point and a gradient so you should be able to get the equation of the normal.
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  5. #5
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    Ahh yes ive done them now, thanks
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  6. #6
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    your welcome !
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  7. #7
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    Quote Originally Posted by Jampop View Post
    Ahh yes ive done them now, thanks
    Hey, if you are going to thank people, click on the "thanks" button! It earns them brownie points.
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