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Math Help - Implicit differentatation?

  1. #1
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    Implicit differentatation?

    I am having difficulty differentiating the following questions

    1. x^(2/3)+y(2/3)=5 This is what I did

    (2/3)x^(-1/3)+(2/3)y^(-1/3) I am also supposed to plot the point(8,1) to find slope


    2. tan(x+y)=x This is what I did

    sec^2(x+y)(1+1)=1 ? (is this correct?)

    3. y=sin(xy)

    1=cos(xy)(1y+1x) is this correct?
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  2. #2
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    Re: Implicit differentatation?

    while differentiating variable y you have to write dy/dx
    as in question 1 (2/3)x^(-1/3) + (2/3)y^(-1/3)dy/dx = 0
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  3. #3
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    Re: Implicit differentatation?

    Quote Originally Posted by waqarhaider View Post
    while differentiating variable y you have to write dy/dx
    as in question 1 (2/3)x^(-1/3) + (2/3)y^(-1/3)dy/dx = 0
    From here, plug in x = 8 and y = 1 to solve for (dy/dx).

    For 3. NO.
    The derivative of y is NOT ONE!!
    The derivative of y is
    dy/dx, or y'.

    You should have y' = cos(xy)*(xy)' -->
    y' = cos(xy)*(xy' + yx'), but x' IS ONE, so we can write this as
    y' = cos(xy)*(xy' + y), or
    dy/dx = cos(xy)*(x*dy/dx + y)
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    Re: Implicit differentatation?

    So would the answer to my second question be

    sec^2(x+y)(1+dy/dx)=1

    Also on a similar query what is meant by d^2y/dx^2 in terms of x and y.

    For a function such as x^2+xy=5

    Does that mean to differentiate two times
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  5. #5
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    Re: Implicit differentatation?

    yes, the derivative of the derivative ...

    \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2}
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    Re: Implicit differentatation?

    tan(x+y)=x

    sec^2(x+y)(1+dy/dx)=1 I am wondering did get the derivative correctly?
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  7. #7
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    Re: Implicit differentatation?

    Quote Originally Posted by homeylova223 View Post
    tan(x+y)=x

    sec^2(x+y)(1+dy/dx)=1 I am wondering did get the derivative correctly?
    so far so good, but get dy/dx by its lonesome ...
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