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Math Help - Implicit Differentiation (iii)

  1. #1
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    Implicit Differentiation (iii)





    What did I do wrong?

    Thanks,
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  2. #2
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    The first derivative is correct \frac{dy}{dx}=\frac{-y^{2}}{x^{2}}

    Now, continue from there by taking the derivative of that. BUT, do not forget to sub in what dy/dx equals. That is where many go astray.

    Quotient rule:

    \frac{x^{2}(-2y)\frac{dy}{dx}-(-y^{2})(2x)}{x^{4}}

    \frac{-2y}{x^{2}}\frac{dy}{dx}+\frac{2y^{2}}{x^{3}}

    Sub in dy/dx from above:

    \frac{-2y}{x^{2}}(\frac{-y^{2}}{x^{2}})+\frac{2y^{2}}{x^{3}}

    \frac{2y^{2}}{x^{3}}+\frac{2y^{3}}{x^{4}}

    y''=\frac{2(x+y)y^{2}}{x^{4}}

    They just used a common denominator and rewrote it. That is the same thing.
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  3. #3
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    Hello, r_maths!

    \frac{1}{x}+\frac{1}{y} \:=\:\frac{1}{\pi}\qquad\text{Find: }\:\frac{dy}{dx}\,\text{ and }\,\frac{d^2y}{dx^2}


    First
    -\frac{1}{x^2} - \frac{1}{y^2}\!\cdot\!\frac{dy}{dx} \:=\:0\quad\Rightarrow\quad \frac{dy}{dx} \:=\:-\frac{y^2}{x^2}\;\;{\color{blue}\hdots\text{ Right!}}

    Second
    -\frac{1}{x^2} - \frac{1}{y^2}\!\cdot\!\frac{dy}{dx} \:=\:0\;\;{\color{blue}\hdots \text{Why did you start back there?}}
    We have: . \frac{dy}{dx}\:=\:\frac{\text{-}y^2}{x^2}

    Then: . \frac{d^2y}{dx^2} \;=\;\frac{x^2(\text{-}2y)\left(\frac{dy}{dx}\right) -(\text{-}y^2)(2x)}{(x^2)^2} \;=\; \frac{-2x^2y\cdot\frac{dy}{dx} + 2xy^2}{x^4}

    . . . . =\;\frac{2xy\left(y - x\!\cdot\!\frac{dy}{dx}\right)}{x^4} \;=\;\frac{2y\left[y - x\left(\text{-}\frac{y^2}{x^2}\right)\right]}{x^3}

    . . . . = \;\frac{2y\left(y + \frac{y^2}{x}\right)}{x^3} \;=\;\frac{2y(xy + y^2)}{x^4}\;=\;\boxed{\frac{2xy + 2y^3}{x^4}}

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Why did you start back there?
    I don't know... The book only showed that method.
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    Quote Originally Posted by Soroban View Post
    . . . . = \;\frac{2y\left(y + \frac{y^2}{x}\right)}{x^3} \;=\;\frac{2y(xy + y^2)}{x^4}\;=\;\boxed{\frac{2xy + 2y^3}{x^4}}
    By the way, your last line, shouldn't it equal to this?
    \;\boxed{\frac{2xy^2 + 2y^3}{x^4}}
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  6. #6
    Moo
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    Hello,

    Quote Originally Posted by r_maths View Post
    By the way, your last line, shouldn't it equal to this?
    \;\boxed{\frac{2xy^2 + 2y^3}{x^4}}
    Yes, but you could find it by yourself
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  7. #7
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    Quote Originally Posted by Moo View Post
    Hello,



    Yes, but you could find it by yourself
    So Soroban is right and the book answer is wrong?
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  8. #8
    Moo
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    Quote Originally Posted by r_maths View Post
    So Soroban is right and the book answer is wrong?
    I've done the calculus too, and I find y, not y...

    So it's 3 against 1
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