Is this the best route to go in this implicit derivation problem?

Question

I have this equation:

$$e^{\frac{x}{y}} = x - y$$

I seem to be going down the wrong path. Is this right so far?

$$\frac{dy}{dx} = e^{\frac{x}{y}} \cdot \frac{dy}{dx} (x \cdot y^{-1}) = 1 - y'$$

$$e^{\frac{x}{y}} ( x \cdot -y^{-2} \cdot y' + y^{-1}) = 1 - y'$$ $$ -e^{\frac{x}{y}} x \cdot y^{-2} \cdot y' + e^{\frac{x}{y}} y^{-1}) = 1 - y'$$

WolframAlpha has this as

$$y'(x) = \frac{y(e^{x/y}-y)}{xe^{x/y}-y^2}$$

Why did the subtraction sign become an equals sign. – Andrew Li May 10 '18 at 12:58 — Andrew Li, May 10 '18 at 12:58

score 2 · Answer 1 · answered May 10 '18 at 13:10

I don't know why your subtraction sign became an equals sign, but your goal, after implicitly differentiating, is to isolate the $y'$ terms so you can factor and divide:

$$e^{x/y} \left( x \cdot -y^{-2} \cdot y' + y^{-1}\right) = 1 - y'$$ $$e^{x/y} \cdot x \cdot -y^{-2} \cdot y' + e^{x/y} \cdot y^{-1} = 1 - y'$$ $$e^{x/y} \cdot x \cdot -y^{-2} \cdot y' + y' = 1 - e^{x/y} \cdot y^{-1}$$

Then factor:

$$y'\left(e^{x/y} \cdot x \cdot -y^{-2} + 1\right) = 1 - e^{x/y} \cdot y^{-1}$$

And divide. After some simplification, you should reach:

$$y' = {{y - e^{x/y} \over y} \cdot -{y^2 \over e^{x/y} x - y^2}} = \frac{y(e^{x/y}-y)}{xe^{x/y}-y^2}$$

@Jwan622 I wrote everything under common denominators in the numerator and denominator of the larger fraction. — Andrew Li, May 10 '18 at 13:57

cansomeonehelpmeout · Accepted Answer · 2018-05-10T13:22:24.897

Your result $$-e^{\frac{x}{y}}xy^{-2}\color{blue}{y'}+e^{\frac{x}{y}}y^{-1}=1-\color{blue}{y'}$$ is correct.

You can continue and simplify this by collecting all the terms with $y'$ as a factor:

$$\begin{align}\color{blue}{y'}\left (1-e^{\frac{x}{y}}xy^{-2}\right )=1-e^{\frac{x}{y}}y^{-1}\end{align}\\\color{blue}{y'}=\frac{1-e^{\frac{x}{y}}y^{-1}}{1-e^{\frac{x}{y}}xy^{-2}}\frac{y^2}{y^2}=\frac{y(y-e^{\frac{x}{y}})}{y^2-e^{\frac{x}{y}}x}$$

This is the same answer that WolframAlpha gave you, only that they have multiplied numerator and denominator by $-1$.

score 1 · Answer 3 · answered May 10 '18 at 13:28

Why is nobody suggesting logarithmic differentiation ?

First take log base $e$ of both sides to get

$$\frac{x}{y(x)}=\ln (x-y(x))\Rightarrow x=y(x)\cdot \ln (x-y(x)).$$

Now differentiate implicitly:

$$\begin{align} 1&=y\cdot \frac{1}{x-y}\cdot \left(1-\frac{dy}{dx}\right)+\ln(x-y)\cdot \frac{dy}{dx} \\ \Rightarrow 1&=\frac{y}{x-y}-\frac{y}{x-y}\cdot \frac{dy}{dx}+\ln(x-y)\cdot \frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}\left(\ln(x-y)-\frac{y}{x-y}\right)&=\frac{x-2y}{x-y} \\ \Rightarrow \frac{dy}{dx}&=\frac{\frac{x-2y}{x-y}}{\ln(x-y)-\frac{y}{x-y}} \\ \Rightarrow \frac{dy}{dx}&=\frac{x-2y}{(x-y)\ln(x-y)-y}. \end{align}$$

OK, maybe now I know why...

score 0 · Answer 4 · answered May 10 '18 at 13:50

0

We have

$$ f(x,y) = e^{\frac{x}{y}}-x+y=0 $$

so we have

$$ df = f_x dx + f_y dy = 0\Rightarrow \frac{dy}{dx}=-\frac{f_x}{f_y} = \frac{y (x-2 y)}{x^2-x y-y^2} $$

answered May 10 '18 at 13:50

Cesareo

33,252

The Dead Legend · Answer 5 · 2018-05-10T13:33:39.490

-1

Please check your line 2, it should be something like this:

$$e^{\frac{x}{y}}\times(y^{-1}-xy^{-2}y')=1-y'$$

Proceed using this to get the WolframAlpha output

edited May 10 '18 at 13:33

answered May 10 '18 at 12:56

The Dead Legend

4,727

1

The OP has it correct. You have to differentiate $y^{-1}$ in the latter term inside the parentheses. – Andrew Li May 10 '18 at 13:00
@AndrewLi Follow this pathway, you would reach the answer and note $dy/dx=y'$ . PS: I triple checked my calculation and it is right. – The Dead Legend May 10 '18 at 13:02
I understand, but applying the product rule the OP's is correct? – Andrew Li May 10 '18 at 13:04
@AndrewLi yes he is, I saw that later. I still don't understand the downvote though xD – The Dead Legend May 10 '18 at 13:09
1

Because your answer is wrong? You differentiated incorrectly. – Andrew Li May 10 '18 at 13:12
$\frac{d}{dx}(y^{-1})\neq -y^{-1}y'$. – cansomeonehelpmeout May 10 '18 at 13:25
@cansomeonehelpmeout $e^{\frac{x}{y}}\times(y^{-1}-xy^{-1}y')=1-y'$. I simply did $d(\frac{x}{y})/dx$ – The Dead Legend May 10 '18 at 13:31
$\frac{d}{dx}(\frac{x}{y})=y^{-1}-xy^{-2}y'$. – cansomeonehelpmeout May 10 '18 at 13:32
1

@cansomeonehelpmeout acknowledged. thank you – The Dead Legend May 10 '18 at 13:33

score -1 · Answer 6 · answered May 10 '18 at 13:08

$e^{\frac{x}{y}} = x - y$

differentiate wrt $y$ and use the chain rule and Quotient rule on $e^{\frac xy}$ in that order ;

$ e^{\frac xy}\cdot \frac{y-\frac{dy}{dx}x}{y^2}=1-\frac{dy}{dx} $

$e^{\frac xy} y -\frac{dy}{dx}x e^{\frac xy} = y^2-y^2\frac{dy}{dx}$

$\frac{dy}{dx}\big(y^2-x\cdot e^{\frac xy}\big) = y^2-e^{\frac xy}y$

$\frac{dy}{dx} = \frac{y(y-e^{\frac xy})}{y^2-x\cdot e^{\frac xy}}$

Is this the best route to go in this implicit derivation problem?

6 Answers6