Results 1 to 10 of 10

Math Help - With respect to y

  1. #1
    Member
    Joined
    Jul 2009
    Posts
    149

    With respect to y

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

    find derivative with respect to y.holding x constant.


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


    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

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

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

    ???? Suggestions?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    The point here is "holding x constant".

    So treat it as though it were:

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

    or even:

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

    ... which should be an elementary application of chain rule.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2009
    Posts
    112
    Quote Originally Posted by el123 View Post
     \frac{x^4}{1-xy}

    find derivative with respect to y.holding x constant.


     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
    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

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

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

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

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

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

    x^4(1-xy)^{-1}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    Quote Originally Posted by Matt Westwood View Post
    The point here is "holding x constant".

    So treat it as though it were:

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

    or even:

    \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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2009
    Posts
    149
    not meant to be equal , meant to be minus lol
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Sep 2009
    Posts
    112
    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 . . .


    <br />
f_y=\frac{x^5}{(1-xy)^2}<br />

    That is right.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jul 2009
    Posts
    149
    so i got it right in the second equation?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Sep 2009
    Posts
    112
    if:
    <br />
f(x,y) = \frac{x^4}{1-xy}<br />


    then:
    <br />
f_y=\frac{x^5}{(1-xy)^2}<br />

    That solution for the partial y is correct.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Jul 2009
    Posts
    149
    gracias amigos!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. One last Dif with respect to x
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 29th 2011, 07:50 PM
  2. Differentiate with respect to x
    Posted in the Calculus Forum
    Replies: 9
    Last Post: January 29th 2011, 07:38 PM
  3. Integrating with respect to x and y
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 29th 2010, 03:24 PM
  4. G(x) in respect to F(X)
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 18th 2009, 07:23 PM
  5. respect to x
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 2nd 2007, 02:01 PM

Search Tags


/mathhelpforum @mathhelpforum