Results 1 to 8 of 8

Math Help - differentiating inverse functions

  1. #1
    Newbie
    Joined
    Oct 2007
    Posts
    19

    differentiating inverse functions

     z=x sin^{-1}\left( \frac{x}{y} \right)

    determine \left(\frac{\partial z}{\partial x}\right)_{y} and  \left(\frac{\partial z}{\partial y}\right)_{x}

    am I right in thinking that:

    x=ysin\left(\frac{z}{x}\right)

    and if so, where do I go from there? I don't suppose i can do this by diff w.r.t x:

    1=\frac{y}{x}cos\left(\frac{z}{x}\right)\left(\fra  c{\partial z}{\partial x}\right)_{y}

    giving me the rather dodgy looking answer:

    \frac{\partial z}{\partial x})_{y}= \frac{x}{ycos(z/x)}

    and for the other one, by diff w.r.t y:

    0= sin(\frac{z}{x}) + ycos(\frac{z}{x}).\frac{\partial z}{\partial y})_{x}

    \frac{\partial z}{\partial y})_{x}= -tan (\frac{z}{x})/y

    thanks, i hope at least the first step is right....
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by UbikPkd View Post
     z=x sin^{-1}\left( \frac{x}{y} \right)

    determine \left(\frac{\partial z}{\partial x}\right)_{y} and  \left(\frac{\partial z}{\partial y}\right)_{x}

    am I right in thinking that:

    x=ysin\left(\frac{z}{x}\right)

    and if so, where do I go from there? I don't suppose i can do this by diff w.r.t x:

    1=\frac{y}{x}cos\left(\frac{z}{x}\right)\left(\fra  c{\partial z}{\partial x}\right)_{y}

    giving me the rather dodgy looking answer:

    \frac{\partial z}{\partial x})_{y}= \frac{x}{ycos(z/x)}

    and for the other one, by diff w.r.t y:

    0= sin(\frac{z}{x}) + ycos(\frac{z}{x}).\frac{\partial z}{\partial y})_{x}

    \frac{\partial z}{\partial y})_{x}= -tan (\frac{z}{x})/y

    thanks, i hope at least the first step is right....
    i do not understand your notation. what does " \left(\frac{\partial z}{\partial x}\right)_{y}" mean? what is that subscript y for? in any case, i do not think it's necessary to solve for x in the first place. you will not get \frac {\partial x}{\partial [ \mbox{something}]} by doing that
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member JaneBennet's Avatar
    Joined
    Dec 2007
    Posts
    293
    You can use the formula \frac{\mathrm{d}}{\mathrm{d}x}\left(\sin^{-1}{x}\right)=\frac{1}{\sqrt{1-x^2}}. So


    \left(\frac{\partial z}{\partial x}\right)_{y}\ =\ \sin^{-1}{\left(\frac{x}{y}\right)}+x\cdot\frac{1}{\sqrt{  1-\left(\frac{x}{y}\right)^2}}\cdot\left(\frac{1}{y}  \right)

    \color{white}.\hspace{19mm}\color{black}=\ \sin^{-1}{\left(\frac{x}{y}\right)}+\frac{x}{\sqrt{y^2-x^2}}


    Similarly for \left(\frac{\partial z}{\partial y}\right)_{x}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2007
    Posts
    19
    Quote Originally Posted by Jhevon View Post
    i do not understand your notation. what does " \left(\frac{\partial z}{\partial x}\right)_{y}" mean? what is that subscript y for? in any case, i do not think it's necessary to solve for x in the first place. you will not get \frac {\partial x}{\partial [ \mbox{something}]} by doing that

    it just means y is to be held constant
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by UbikPkd View Post
    it just means y is to be held constant
    ok, that is weird though. i've never seen that notation. usually you would just write \frac {\partial z}{\partial x} and it is understood that all variables beside x are treated as constants. anyway, JaneBennet assumed that's what was meant, so follow her post.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,961
    Thanks
    349
    Awards
    1
    Quote Originally Posted by Jhevon View Post
    ok, that is weird though. i've never seen that notation. usually you would just write \frac {\partial z}{\partial x} and it is understood that all variables beside x are treated as constants. anyway, JaneBennet assumed that's what was meant, so follow her post.
    I don't know how often it appears in pure Math, but the notation is common in Thermodynamics. Functions are typically given in terms of three "extensive" variables such as volume, temperature, and pressure. (V, T, and P). The derivative \left ( \frac{\partial U}{\partial T} \right ) _P is the derivative of the internal energy function U(P, V, T)with respect to the temperature at constant pressure. The volume is allowed to change as a function of T in this case, so technically this isn't a partial derivative at all though the derivative operation on the function U is calculated as if it were a partial derivative.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2007
    Posts
    19
    Quote Originally Posted by JaneBennet View Post
    You can use the formula \frac{\mathrm{d}}{\mathrm{d}x}\left(\sin^{-1}{x}\right)=\frac{1}{\sqrt{1-x^2}}. So


    \left(\frac{\partial z}{\partial x}\right)_{y}\ =\ \sin^{-1}{\left(\frac{x}{y}\right)}+x\cdot\frac{1}{\sqrt{  1-\left(\frac{x}{y}\right)^2}}\cdot\left(\frac{1}{y}  \right)

    \color{white}.\hspace{19mm}\color{black}=\ \sin^{-1}{\left(\frac{x}{y}\right)}+\frac{x}{\sqrt{y^2-x^2}}


    Similarly for \left(\frac{\partial z}{\partial y}\right)_{x}.
    yer i like this, thanks! I'll have to prove that formula (i can't just quote it) but i've done that so the first one is done. However, for \left(\frac{\partial z}{\partial y}\right)_{x} i don't see how i can use the same formula as the variable being differentiated, y, is now on the bottom, so isn't a different one is needed?

    ie
    \frac{\mathrm{d}}{\mathrm{d}y}\left(\sin^{-1}{(1/y)}\right)=?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by UbikPkd View Post
    yer i like this, thanks! I'll have to prove that formula (i can't just quote it)
    it's a well-known rule, you can just use it. i very much doubt you would be required to prove it


    but i've done that so the first one is done. However, for \left(\frac{\partial z}{\partial y}\right)_{x} i don't see how i can use the same formula as the variable being differentiated, y, is now on the bottom, so isn't a different one is needed?

    ie
    \frac{\mathrm{d}}{\mathrm{d}y}\left(\sin^{-1}{(1/y)}\right)=?
    use the chain rule, as JaneBennet did.

    \frac {\partial }{\partial y} \sin^{-1} \left( \frac xy \right) = \frac 1{\sqrt{1 - \left( \frac xy \right)^2}} \cdot \left( \frac {\partial }{\partial y} ~\frac xy \right)

    so what is \frac {\partial }{\partial y}~\frac xy?

    note: you will not need the product rule here, since x is considered a constant
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Differentiating some functions.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 17th 2011, 04:07 AM
  2. Differentiating sin functions
    Posted in the Calculus Forum
    Replies: 6
    Last Post: March 27th 2011, 04:16 PM
  3. Differentiating Inverse Functions
    Posted in the Calculus Forum
    Replies: 7
    Last Post: March 22nd 2011, 12:47 AM
  4. Differentiating the inverse
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 6th 2009, 04:04 PM
  5. Differentiating the Inverse
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 7th 2008, 03:54 PM

Search Tags


/mathhelpforum @mathhelpforum