# implicit differentiation of exponentials containing x and y

So I'm trying to differentiate an equation using implicit differentiation.

I start with $e^{x/y} = 7x - y$

Now the left side of the eqn is where I'm having trouble.

I tried to use differentiation rules for exponentials, but this is incorrect.

Here's what I tried though:

$$(e^x)^{1/y} \ln(e^x) y' = 7 - y'$$

simplfied to:

$$x(e^x)^{1/y} \cdot y' = RS$$

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Try taking $\ln$ of both sides. –  Baby Dragon Jul 8 '13 at 4:06

## migrated from mathoverflow.netJul 8 '13 at 3:55

This question came from our site for professional mathematicians.

I think the key thing to remember is: $y$ is a function of $x$, so $xy^{-1}$ is a function of $x$, call it $u(x)$: $u(x) = xy^{-1}$. Now the derivative of $e^{u(x)}$ is $e^{u(x)}u'(x)$, so we need $u'(x) = (xy^{-1})' = y^{-1} - xy^{-2}y'$. Thus

$e^{xy^{-1}}(y^{-1} - xy^{-2}y') = 7 - y'$,

an equation which is linear in $y'$, for which we can solve using some simple algebra:

$(1 - e^{xy^{-1}}xy^{-2})y' = 7 - e^{xy^{-1}}y^{-1}$,

or

$y' = \frac{(7 - e^{xy^{-1}}y^{-1})}{(1- e^{xy^{-1}}xy^{-2})} = \frac{(7y^2 - ye^{xy^{-1}})}{(y^2- xe^{xy^{-1}})}$,

which is about as far as we can go without knowing $y(x)$. Of course it should be remembered that, in deriving this formula, we have assumed that $y(x) \ne 0$ is a differentiable function of $x$.

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I deleted and thne re-posted this answer since I accidentally made the previous vs. community wiki, and couldn't figure out how to turn CW off. Sorry, thanks, and enjoy the post! –  1950Robert Lewis Jul 8 '13 at 17:36