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Math Help - Implicit Differentiation

  1. #1
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    Implicit Differentiation

    How can I 'evaluate' the following derivative and simplify it ?

    xy^2 = 3^(y/ln x) + sin^2 x
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  2. #2
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    I assume you are taking the derivative with respect to x? If so you need to use the chain rule, i.e. \frac{d}{dx} (f\circ g)(x) = (f'\circ g)(x)\cdot g'(x).

    To start you off
    \frac{d}{dx}\left[xy^2\right] = y^2 + x\cdot 2y\cdot y'.

    Also, if f(x) = a^{g(x)} then
    \frac{d}{dx} f(x) = \frac{d}{dx} a^{g(x)} \Rightarrow \frac{d}{dx} \left[\ln f(x)\right] = \frac{d}{dx} \left[g(x)\cdot \ln a\right] \Rightarrow \frac{1}{f(x)}\cdot f'(x) = \ln a \cdot g'(x) \Rightarrow f'(x) = f(x) \cdot g'(x) \ln a

    I hope that helps.
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  3. #3
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by lvleph View Post

    \frac{d}{dx}\left[xy^2\right] = y^2 + x\cdot 2y\cdot y'.

    \displaystyle  \frac {d}{d x } (xy^2) = y^2

    you don't need to do it like that... (i assume that your last " y' " is derivated by the x ) it's much easier (for more complex problems) to just get the constant (in this case y) out ...

    \displaystyle   \frac {d}{d x } (xy^2) = y^2  \cdot [ \frac {d}{d x } (x) ]= y^2


    this :

     (u\cdot v)' = u'v +v'u

    is being used for derivate something like this (where are two functions )

    \displaystyle   \frac {d}{d x } (e^x \cdot x^2 )

    that what you did there is like that for (2x^2) for derivate on "x" you do  (2x^2)'= 2' \cdot x^2 + 2 \cdot (x^2)' = 0\cdot x^2 + 2\cdot 2x = 4x that's to much work to be done, especially when we have constant on our hands

    sorry i don't have intention to be rude or something but to do derivation of function and constant like derivation of two functions ... that's not wrong but not needed at all
    Last edited by yeKciM; August 24th 2010 at 09:46 AM.
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  4. #4
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    Quote Originally Posted by yeKciM View Post
    \displaystyle  \frac {\delta}{\delta x } (xy^2) = y^2

    you don't need to do it like that... (i assume that your last " y' " is derivated by the x ) it's much easier (for more complex problems) to just get the constant (in this case y) out ...

    \displaystyle   \frac {\delta}{\delta x } (xy^2) = y^2 \cdot \frac {\delta}{\delta x } (x) = y^2
    Except that you are assuming that y is a constant with respect to x and I was not assuming this. Because there was no statement of dependence of y on x it is more correct to assume dependence and then if there is no dependence we can just say y'=0 and everything simplifies to your case.
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  5. #5
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by lvleph View Post
    Except that you are assuming that y is a constant with respect to x and I was not assuming this. Because there was no statement of dependence of y on x it is more correct to assume dependence and then if there is no dependence we can just say y'=0 and everything simplifies to your case.
    yes but everything that is not "x" when you do derivation on x is constant
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  6. #6
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    No, it is not. Here is a good example. \frac{d}{dx}\left[x\cdot f^2(x)\right] = x\cdot2f(x) \cdot f'(x) + f(x).
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  7. #7
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by lvleph View Post
    No, it is not. Here is a good example. \frac{d}{dx}\left[x\cdot f^2(x)\right] = x\cdot2f(x) \cdot f'(x) + f(x).
    yes but, that (f(x)) is function (or any mapping) of x. or am I mistaking ? it's not like function of y or anything else

    sorry, but now i'm like confused



    Edit: aaaah sorry I totally forgot about question up there... I'm very very sorry
    Last edited by yeKciM; August 24th 2010 at 10:17 AM.
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  8. #8
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    Quote Originally Posted by Sangreal View Post
    How can I 'evaluate' the following derivative and simplify it ?

    xy^2 = 3^(y/ln x) + sin^2 x
    Differentiating each term independently...

    LHS

    xy^2\;\;\; is a product.

    \displaystyle\frac{d}{dx}\left(xy^2\right)=y^2\fra  c{d}{dx}x+x\frac{d}{dx}{y^2}=y^2+x\frac{dy}{dx}\ \frac{d}{dy}y^2

    =\displaystyle\ y^2+2xy\frac{dy}{dx}


    RHS

    \displaystyle\huge\ 3^{\left(\frac{y}{lnx}\right)}=3^{h(x)}

    \displaystyle\frac{d}{dx}3^{h(x)}=\frac{d}{dh(x)}3  ^{h(x)}\frac{d}{dx}h(x)=3^{h(x)}ln3\frac{d}{dx}h(x  )

    since \displaystyle\frac{d}{dx}a^x=a^xlna

    giving

    \displaystyle\frac{d}{dx}3^{\left(\frac{y}{lnx}\ri  ght)}=3^{\left(\frac{y}{lnx}\right)}ln3\frac{d}{dx  }\left(\frac{y}{lnx}\right)

    \displaystyle\frac{d}{dx}\left(\frac{y}{lnx}\right  )=\frac{lnx\frac{dy}{dx}-y\left(\frac{1}{x}\right)}{(lnx)^2}


    Differentiating Sin^2(x)

    let u=Sin(x)

    \displaystyle\frac{d}{du}u^2\frac{du}{dx}=2uCosx=2  SinxCosx=Sin(2x)

    You need to get \displaystyle\frac{dy}{dx}=........ as \displaystyle\frac{dy}{dx} occurs more than one when you write LHS=RHS.

    so you will need to factor it out...
    Last edited by Archie Meade; August 24th 2010 at 11:45 AM. Reason: small typo
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