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Thread: With respect to y

  1. #1
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    With respect to y

    $\displaystyle \frac{x^4}{1-xy}$

    find derivative with respect to y.holding x constant.


    $\displaystyle f_y=\frac{[(4x^3)(1-xy)]-[(x^4)(-x)]}{(1-xy)^2}$


    $\displaystyle f_y= \frac{4x^3-4x^4y+x^5}{(1-xy)^2}$

    Is that right or did i do that wrong?
    Or is is suppose to be

    $\displaystyle f_y=\frac{[(0)(1-xy)]-[(x^4)(-x)]}{(1-xy)^2}$

    $\displaystyle f_y=\frac{x^5}{(1-xy)^2}$

    ???? Suggestions?
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  2. #2
    MHF Contributor Matt Westwood's Avatar
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    The point here is "holding x constant".

    So treat it as though it were:

    $\displaystyle \frac d {dy} \left({\frac {c^4}{1-cy}}\right)$

    or even:

    $\displaystyle \frac d {dy} \left({\frac {a}{1-by}}\right)$

    ... which should be an elementary application of chain rule.
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  3. #3
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    Quote Originally Posted by el123 View Post
    $\displaystyle \frac{x^4}{1-xy}$

    find derivative with respect to y.holding x constant.


    $\displaystyle f_y=\frac{[(4x^3)(1-xy)]=[(x^4)(-x)]}{(1-xy)^2}$
    Why is there an equal sign here?

    Quote Originally Posted by el123 View Post
    $\displaystyle f_y= \frac{4x^3-4x^4y+x^5}{(1-xy)^2}$

    Is that right or did i do that wrong?
    Or is is suppose to be

    $\displaystyle f_y=\frac{[(0)(1-xy)]=[(x^4)(-x)]}{(1-xy)^2}$

    $\displaystyle f_y=\frac{x^5}{(1-xy)^2}$

    ???? Suggestions?
    Is that supposed to be the partial y for your very first expression?

    $\displaystyle \frac{x^4}{1-xy}$

    because if you consider y as the only variable it's just power rule.

    $\displaystyle x^4(1-xy)^{-1}$
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  4. #4
    MHF Contributor Matt Westwood's Avatar
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    Quote Originally Posted by Matt Westwood View Post
    The point here is "holding x constant".

    So treat it as though it were:

    $\displaystyle \frac d {dy} \left({\frac {c^4}{1-cy}}\right)$

    or even:

    $\displaystyle \frac d {dy} \left({\frac {a}{1-by}}\right)$

    ... which should be an elementary application of chain rule.
    In fact, in the last bit of your post I see you've done it right.
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  5. #5
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    not meant to be equal , meant to be minus lol
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  6. #6
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    Quote Originally Posted by Matt Westwood View Post
    In fact, in the last bit of your post I see you've done it right.
    oh hmm you're right I didn't notice that either . . .


    $\displaystyle
    f_y=\frac{x^5}{(1-xy)^2}
    $

    That is right.
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  7. #7
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    so i got it right in the second equation?
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  8. #8
    MHF Contributor Matt Westwood's Avatar
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    Quote Originally Posted by el123 View Post
    so i got it right in the second equation?
    Yes - good job. Partial differentiation, it's more simple than you'd initially suspect.
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  9. #9
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    if:
    $\displaystyle
    f(x,y) = \frac{x^4}{1-xy}
    $


    then:
    $\displaystyle
    f_y=\frac{x^5}{(1-xy)^2}
    $

    That solution for the partial y is correct.
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  10. #10
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    gracias amigos!
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