Differentiate $e^{y/x} = 20x-y$ I am trying to use implicit differentiation to differentiate $e^{y/x} = 20x-y$. I get $\frac{20}{2 e^{y/x} \cdot \frac{x-y}{x^2}}$, but according to the math website I'm using, "WebWork", this is wrong. 
I'm not sure how to handle it when an equation has two $\frac{dy}{dx}$ floating around, but I'm not sure this is the problem.
Here are my steps:
$$\frac{d}{dx}(e^{y/x}) = \frac{d}{dx}(20x-y)$$
Factoring both sides independently:
$$\frac{d}{dx}(e^{y/x}) = e^{y/x}\cdot\frac{x-y}{x^2} \frac{dy}{dx}$$
$$\frac{d}{dx}(20x-y)=20-\frac{dy}{dx}$$
Finding $\frac{dy}{dx}$
$$e^{y/x} \cdot \frac{x-y}{x^2} \frac{dy}{dx} = 20 - \frac{dy}{dx}$$
$$\frac{dy}{dx}+\frac{x-y}{x^2} \frac{dy}{dx} = \frac{20}{e^{y/x}}$$
$$\frac{dy}{dx} + \frac{dy}{dx} = \frac{20}{e^{y/x}\cdot\frac{x-y}{x^2}}$$
$$2\frac{dy}{dx} = \frac{20}{e^{y/x}\cdot\frac{x-y}{x^2}}$$
$$\frac{dy}{dx} = \frac{20}{2e^{y/x}\cdot\frac{x-y}{x^2}}$$
 A: You get:
$$(e^{y/x})'_x=(2x-y)'_x \Rightarrow \\
e^{y/x}\cdot (y/x)'_x=2-y'\Rightarrow \\
(2x-y)\cdot \frac{y'x-y}{x^2}=2-y' \Rightarrow \\
(2x^2-xy)y'-(2x-y)y=2x^2-x^2y'\Rightarrow \\
y'=\frac{2x^2+2xy-y^2}{3x^2-xy}.$$
If you are familiar with multivariable calculus, consider: $F(x,y)=e^{y/x}-2x+y=0$. Then:
$$y'=-\frac{F_x}{F_y}=-\frac{e^{y/x}\cdot (-y/x^2)-2}{e^{y/x}\cdot (1/x)+1}=\frac{(2x-y)y+2x^2}{(2x-y)x+x^2}=\frac{2x^2+2xy-y^2}{3x^2-xy}.$$
A: Consider representing $\frac{dy}{dx}=y'$ to make the equation clearer and easier to deal with terms.
Assuming your original equation is $e^{y/x} = 20x-y$.

$\dfrac{d}{dx}(e^{y/x}) = e^{y/x}\cdot\dfrac{x-y}{x^2} \dfrac{dy}{dx}.\quad$ Incorrect! You have to consider $y$ as a function of $x$ as well.

$\dfrac{d}{dx}(e^{y/x})=e^{y/x}\cdot \dfrac{d}{dx}\left(\dfrac{y}x\right)=e^{y/x}\cdot\dfrac{x\dfrac{dy}{dx}-y\dfrac{dx}{dx}}{x^2}=e^{y/x}\cdot\left(\dfrac{xy'-y}{x^2}\right)\tag1$
So $\dfrac{d}{dx}\left(e^{y/x} = 20x-y\right)\implies\dfrac{d}{dx}\left(e^{y/x} \right)=\dfrac{d}{dx}\left(20x-y\right)$. Upon substituting $(1)$ we have,
$\begin{align}e^{y/x}\cdot\left(\dfrac{xy'-y}{x^2}\right)=20-y'&\implies e^{y/x}xy'-e^{y/x}y=20x^2-x^2y'\\&\implies y'(x^2+xe^{y/x})=20x^2+e^{y/x}y\end{align}$
I suppose you can take it from here.
Taking logarithms can be considered as well.$$\frac{y}x=\ln(20x-y)$$ Differentiating $$\frac{xy'-y}{x^2}=\frac{20-y'}{20x-y} \implies y'=\frac{20x^2+20xy-y^2}{21x^2-xy}$$
A: $e^{y/x} \cdot \frac{x-y}{x^2} = 20 - \frac{dy}{dx}$. You forgot that there was a $\frac{dy}{dx}$ term on the left which should have prevented you from simply adding those two fractions together unless you had slid it over into the numerator. But I don't even see it there. Compare your solution to mine:
$$
e^{\frac{y}{x}} = 20x-y\\
\frac{d}{dx}\left(e^{\frac{y}{x}}\right) = \frac{d}{dx}(20x-y)\\
e^{\frac{y}{x}}\frac{d}{dx}\left(\frac{y}{x}\right)=20-\frac{dy}{dx}\\
e^{\frac{y}{x}}\left(\frac{d}{dx}\left(\frac{1}{x}\right)y+\frac{1}{x}\frac{dy}{dx}\right)=20-\frac{dy}{dx}\\
e^{\frac{y}{x}}\left(-\frac{1}{x^2}y+\frac{1}{x}\frac{dy}{dx}\right)=20-\frac{dy}{dx}\\
-\frac{1}{x^2}ye^{\frac{y}{x}}+\frac{1}{x}\frac{dy}{dx}e^{\frac{y}{x}}=20-\frac{dy}{dx}\\
\frac{1}{x}\frac{dy}{dx}e^{\frac{y}{x}}+\frac{dy}{dx}=20+\frac{1}{x^2}ye^{\frac{y}{x}}\\
\frac{dy}{dx}\left(\frac{1}{x}e^{\frac{y}{x}}+1\right)=20+\frac{1}{x^2}ye^{\frac{y}{x}}\\
\frac{dy}{dx}=\frac{20+\frac{1}{x^2}ye^{\frac{y}{x}}}{\frac{1}{x}e^{\frac{y}{x}}+1}\\
\frac{dy}{dx}=\frac{20+\frac{1}{x^2}ye^{\frac{y}{x}}}{\frac{1}{x}e^{\frac{y}{x}}+1}\cdot\frac{x^2}{x^2}\\
\frac{dy}{dx}=\frac{20x^2+ye^{\frac{y}{x}}}{xe^{\frac{y}{x}}+x^2}\\
$$
Wolfram Alpha check.
