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Math Help - Implicit differetiation

  1. #1
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    Implicit differetiation

    Please show me how you work this out. Thanks
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  2. #2
    Moo
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    Hello,

    \boxed{e^{xy} \sin(yz)-e^{yz}\cos(xz)+z=0}

    --> ?=\frac{\partial z}{\partial y}=z'(y) (for more comfort)

    ---> \boxed{e^{x{\color{red}y}} \sin({\color{red}y}{\color{blue}z})-e^{{\color{red}y}{\color{blue}z}}\cos(x{\color{blu  e}z})+{\color{blue}z}=0}


    ~~~~~~~~~~
    The first term will use the product rule since there are y and z in it, considering x as a constant and y as the variable.

    Thus the derivative of the first term will be :

    xe^{xy} \sin(yz)+e^{xy} {\color{red}\dots} \cos(yz)

    The red dots correspond to the derivative of yz. We have to use the product rule.
    ---> {\color{red}\dots}=z+yz'

    ---------> \boxed{xe^{xy} \sin(yz)+e^{xy} (z+yz') \cos(yz)} \ (1)


    ~~~~~~~~~~
    Now considering the second term e^{{\color{red}y}{\color{blue}z}}\cos(x{\color{blu  e}z})

    We have to use the product rule too.

    \underbrace{\color{red}\dots}_{\text{derivative of yz, again}} e^{yz} \cos(xz) \ \ + \ \ e^{yz} (-(xz') \sin(xz))

    ---------> \boxed{(z+yz') e^{yz} \cos(xz) - xz' e^{yz} \sin(xz)} \ (2)


    ~~~~~~~~~~
    Now, the third term, z.
    Its derivative is z' (3)



    ------------------------------

    The derivative of the whole thing will be (1)-(2)+(3)=0

    And... I won't copy it, you can do it

    I hope there is no mistake, check it again, it will make a good exercise
    Last edited by Moo; May 2nd 2008 at 02:18 PM.
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  3. #3
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    this is also what I need, thax
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  4. #4
    Moo
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    Quote Originally Posted by vincent8818 View Post
    this is also what I need, thax
    What didn't you get ?
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  5. #5
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    z+yz'??
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  6. #6
    Moo
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    Quote Originally Posted by vincent8818 View Post
    z+yz'??
    Ok,

    This is the derivative of yz, with respect to y.

    It's like having y*z(y) (z : function of y).

    Thus applying the product rule, we get :

    1*z(y)+y*z'(y)

    1 is the derivative of y, with respect to y.

    So it's z(y)+yz'(y), which I simplified into z+yz'
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  7. #7
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    now i got it, thax
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  8. #8
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    Quote Originally Posted by Moo View Post
    Hello,

    \boxed{e^{xy} \sin(yz)-e^{yz}\cos(xz)+z=0}

    --> ?=\frac{\partial z}{\partial y}=z'(y) (for more comfort)

    ---> \boxed{e^{x{\color{red}y}} \sin({\color{red}y}{\color{blue}z})-e^{{\color{red}y}{\color{blue}z}}\cos(x{\color{blu  e}z})+{\color{blue}z}=0}


    ~~~~~~~~~~
    The first term will use the product rule since there are y and z in it, considering x as a constant and y as the variable.

    Thus the derivative of the first term will be :

    xe^{xy} \sin(yz)+e^{xy} {\color{red}\dots} \cos(xy)

    The red dots correspond to the derivative of yz. We have to use the product rule.
    ---> {\color{red}\dots}=z+yz'

    ---------> \boxed{xe^{xy} \sin(yz)+e^{xy} (z+yz') \cos(xy)} \ (1)


    ~~~~~~~~~~
    Now considering the second term e^{{\color{red}y}{\color{blue}z}}\cos(x{\color{blu  e}z})

    We have to use the product rule too.

    \underbrace{\color{red}\dots}_{\text{derivative of yz, again}} e^{yz} \cos(xz) \ \ + \ \ e^{yz} (-(xz') \sin(xz))

    ---------> \boxed{(z+yz') e^{yz} \cos(xz) - xz' e^{yz} \sin(xz)} \ (2)


    ~~~~~~~~~~
    Now, the third term, z.
    Its derivative is z' (3)



    ------------------------------

    The derivative of the whole thing will be (1)-(2)+(3)=0

    And... I won't copy it, you can do it

    I hope there is no mistake, check it again, it will make a good exercise
    impressive!
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