differentiating inverse functions

**UbikPkd** · December 23rd 2007, 08:41 AM

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

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....

**Jhevon** · December 23rd 2007, 09:04 AM

Originally Posted by UbikPkd

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

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

**JaneBennet** · December 23rd 2007, 09:18 AM

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 $\left(\frac{\partial z}{\partial y}\right)_{x}$ .

**UbikPkd** · December 23rd 2007, 10:45 AM

Originally Posted by Jhevon

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

**Jhevon** · December 23rd 2007, 10:52 AM

Originally Posted by UbikPkd

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.

**topsquark** · December 23rd 2007, 12:15 PM

Originally Posted by Jhevon

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

**UbikPkd** · December 23rd 2007, 01:58 PM

Originally Posted by JaneBennet

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 $\frac{\mathrm{d}}{\mathrm{d}y}\left(\sin^{-1}{(1/y)}\right)=?$

**Jhevon** · December 23rd 2007, 02:04 PM

Originally Posted by UbikPkd

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

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