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    derivative very tricky : need guide

    How are these pattered in the rule of derivatives?
    Find the derivative of dy/dx of the relations:

    1.) x^3y+ 2y^4 - x^4 = 2xy


    and

    2.) x^2 = x+2y/ x - 2y




    thanks a lot
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    Re: derivative very tricky : need guide

    If I want to simplify \frac{d(y)}{dx}, I obviously can't do it that easily as the variables are wrong.

    However, I can apply the chain rule:  \frac{d(y)}{dx}.\frac{dx}{dy}=\frac{d(y)}{dy}

    Solving for \frac{d(y)}{dx}, I get \frac{d(y)}{dx}=\frac{d(y)}{dy}.\frac{dy}{dx}

    Evaluate \frac{d(y)}{dy} which is 1. Therefore \frac{d(y)}{dx}= \frac{dy}{dx}

    Now try apply these principles in your question
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    Re: derivative very tricky : need guide

    ill try sir... ill be back and let you see my answer also... hope i can get it right

    thanks
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    Re: derivative very tricky : need guide

    Quote Originally Posted by jgv115 View Post
    If I want to simplify \frac{d(y)}{dx}, I obviously can't do it that easily as the variables are wrong.

    However, I can apply the chain rule:  \frac{d(y)}{dx}.\frac{dx}{dy}=\frac{d(y)}{dy}

    Solving for \frac{d(y)}{dx}, I get \frac{d(y)}{dx}=\frac{d(y)}{dy}.\frac{dy}{dx}

    Evaluate \frac{d(y)}{dy} which is 1. Therefore \frac{d(y)}{dx}= \frac{dy}{dx}

    Now try apply these principles in your question
    To be honest, I don't understand this. What is \frac{d(y)}{dx} and how is it different from \frac{dy}{dx}?

    OP, it seems that you need to differentiate an implicit function. Obviously, you need to read how to do this, for example, in Wikipedia. Once you learn the method and read several finished examples, please describe the exact difficulty you are having with this particular problem. The thing is that this forum is not a tutorial service and you can't expect people to explain completely new material to you. The forum is here to help those who know the material but struggle to apply it in a concrete situation.
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    Re: derivative very tricky : need guide

    thank you
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    Re: derivative very tricky : need guide

    i'll give you an example, perhaps you'll see how it works:

    if x2+y2 = 1, find dy/dx.

    step 1) differentiate both sides with respect to x:

    2x(dx/dx) + 2y(dy/dx) = 0 but dx/dx = 1, so:

    2x + 2y(dy/dx) = 0

    step 2) solve for dy/dx:

    2y(dy/dx) = -2x

    dy/dx = -2x/2y = -x/y
    Thanks from HallsofIvy
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    Re: derivative very tricky : need guide

    So this is called implicit differentiation. You can differentiate both sides with respect to x where y is a function of x and \frac{dx}{dx} = 1:

     \begin{aligned} x^3 y + 2 y^4 -x^4 & = 2xy \\ x^3 \frac{dy}{dx} + 3x^2 y + 8y^3 \frac{dy}{dx} - 4x^3 & = 2y + 2x \frac{dy}{dx} \\  \\ [\text{ Now you gather dy/dx one side and isolating dy/dx }] \\ \\ \frac{dy}{dx} & = \frac{2y  - 3x^2 y + 4 x^3}{x^3 + 8 y^3 - 2x} \end{aligned}

    For the second one:

     \begin{aligned} x^2 & =  x + \frac{2y}{x} - 2y \\ 2x & = 1 + \frac{x \frac{2 dy}{dx} - 2y}{x^2} - 2 \frac{dy}{dx} \\ \\ [\text{Same as before gather dy/dx together and simplifying using Algebra you get }] \\ \\ \frac{dy}{dx} & = \frac{2x^3 -x^2 + 2y}{2x - 2x^2} \end{array}

    You can check it to confirm here for the first equation:

    differentiation x^3 y +2 y^4 -x^4 - 2xy = 0 - Wolfram|Alpha

    And for the second one:

    differentiation x^2 - x - 2y/x +2y = 0 - Wolfram|Alpha

    Hope this helps.
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    Re: derivative very tricky : need guide

    a million of thanks for the solution sir... the best Helper of this site... --- rcs ----
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    Re: derivative very tricky : need guide

    Quote Originally Posted by rcs View Post
    a million of thanks for the solution sir... the best Helper of this site... --- rcs ----
    You're welcome. Glad to help.
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    Re: derivative very tricky : need guide

    Quote Originally Posted by x3bnm View Post
    So this is called implicit differentiation. You can differentiate both sides with respect to x where y is a function of x and \frac{dx}{dx} = 1:

     \begin{aligned} x^3 y + 2 y^4 -x^4 & = 2xy \\ x^3 \frac{dy}{dx} + 3x^2 y + 8y^3 \frac{dy}{dx} - 4x^3 & = 2y + 2x \frac{dy}{dx} \\  \\ [\text{ Now you gather dy/dx one side and isolating dy/dx }] \\ \\ \frac{dy}{dx} & = \frac{2y  - 3x^2 y + 4 x^3}{x^3 + 8 y^3 - 2x} \end{aligned}



     \begin{aligned} x^2 & =  x + \frac{2y}{x} - 2y \\ 2x & = 1 + \frac{x \frac{2 dy}{dx} - 2y}{x^2} - 2 \frac{dy}{dx} \\ \\ [\text{Same as before gather dy/dx together and simplifying using Algebra you get }] \\ \\ \frac{dy}{dx} & = \frac{2x^3 -x^2 + 2y}{2x - 2x^2} \end{array}

    You can check it to confirm here for the first equation:

    differentiation x^3 y +2 y^4 -x^4 - 2xy = 0 - Wolfram|Alpha

    And for the second one:

    differentiation x^2 - x - 2y/x +2y = 0 - Wolfram|Alpha

    Hope this helps.
    i found it confusing .. how did you get this
    For the second one:

    x^2 = 1 + 2y/x - 2y sir how did you this ?

    because it is supposed to be in number 2: the equation is x^2 = (x+2y)/(x-2y)

    thanks
    Last edited by rcs; September 2nd 2012 at 06:42 PM.
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    Re: derivative very tricky : need guide

    x3bnm read the question AS YOU WROTE IT. If you had used the brackets in the correct place in the first place, you would have been given a correct solution.

    Anyway, you now know how to differentiate implicitly, so I suggest you try it out for yourself with the correct question.
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    Re: derivative very tricky : need guide

    Quote Originally Posted by Prove It View Post
    x3bnm read the question AS YOU WROTE IT. If you had used the brackets in the correct place in the first place, you would have been given a correct solution.

    Anyway, you now know how to differentiate implicitly, so I suggest you try it out for yourself with the correct question.
    is it possible to get the derivative like ,
    x^2(x-2y) = x + 2y sir?
    multiplying both side by x - 2y of the equation, before taking the derivative
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    Re: derivative very tricky : need guide

    Quote Originally Posted by rcs View Post
    is it possible to get the derivative like ,
    x^2(x-2y) = x + 2y sir?
    multiplying both side by x - 2y of the equation, before taking the derivative
    Yes the new solution is given below:

    \begin{align*} x^2 =& \frac{(x+2y)}{(x-2y)} \\  2x =& \frac{(x - 2y)(1 + 2 \frac{dy}{dx}) - (x+2y)(1 - 2 \frac{dy}{dx})}{(x-2y)^2} \\  2x =& \frac{x + 2x\frac{dy}{dx} - 2y -4y \frac{dy}{dx} -x + 2x\frac{dy}{dx} -2y +4y\frac{dy}{dx}}{(x -2y)^2} \\  2x(x - 2y)^2 =&  x + 2x\frac{dy}{dx} - 2y -4y \frac{dy}{dx} -x + 2x\frac{dy}{dx} -2y +4y\frac{dy}{dx} \\  2x(x - 2y)^2 + 4y =& 4x \frac{dy}{dx} \\  \frac{dy}{dx} =& \frac{2x(x - 2y)^2 + 4y}{4x} \\  \frac{dy}{dx} =& \frac{x(x-2y)^2 + 2y}{2x} \\ \frac{dy}{dx} =& \frac{x^3 - 4x^2 y + 4 xy^2 + 2y}{2x} \text{.................[after simplification]  } \end{align*}

    Hope it will help you.

    Also you can check the answer at:

    differentiation x^2 = (x+2y)/(x-2y) - Wolfram|Alpha
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    Re: derivative very tricky : need guide

    Quote Originally Posted by x3bnm View Post
    Yes the new solution is given below:

    \begin{align*} x^2 =& \frac{(x+2y)}{(x-2y)} \\  2x =& \frac{(x - 2y)(1 + 2 \frac{dy}{dx}) - (x+2y)(1 - 2 \frac{dy}{dx})}{(x-2y)^2} \\  2x =& \frac{x + 2x\frac{dy}{dx} - 2y -4y \frac{dy}{dx} -x + 2x\frac{dy}{dx} -2y +4y\frac{dy}{dx}}{(x -2y)^2} \\  2x(x - 2y)^2 =&  x + 2x\frac{dy}{dx} - 2y -4y \frac{dy}{dx} -x + 2x\frac{dy}{dx} -2y +4y\frac{dy}{dx} \\  2x(x - 2y)^2 + 4y =& 4x \frac{dy}{dx} \\  \frac{dy}{dx} =& \frac{2x(x - 2y)^2 + 4y}{4x} \\  \frac{dy}{dx} =& \frac{x(x-2y)^2 + 2y}{2x} \\ \frac{dy}{dx} =& \frac{x^3 - 4x^2 y + 4 xy^2 + 2y}{2x} \text{.................[after simplification]  } \end{align*}

    Hope it will help you.

    Also you can check the answer at:

    differentiation x^2 = (x+2y)/(x-2y) - Wolfram|Alpha
    thanks x3bmn you are also nice and kind. God BLess
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    Re: derivative very tricky : need guide

    Such a big help sir. Thanks
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