Differentiate $e^{y/x} = 20x-y$

Question

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 $\frac{dy}{dx}$

$$e^{y/x} \cdot \frac{x-y}{x^2} \frac{dy}{dx} = 20 - \frac{dy}{dx}$$

$$\frac{dy}{dx}+\frac{x-y}{x^2} \frac{dy}{dx} = \frac{20}{e^{y/x}}$$

$$\frac{dy}{dx} + \frac{dy}{dx} = \frac{20}{e^{y/x}\cdot\frac{x-y}{x^2}}$$

$$2\frac{dy}{dx} = \frac{20}{e^{y/x}\cdot\frac{x-y}{x^2}}$$

$$\frac{dy}{dx} = \frac{20}{2e^{y/x}\cdot\frac{x-y}{x^2}}$$

Answer 1

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 $$\frac{xy'-y}{x^2}=\frac{20-y'}{20x-y} \implies y'=\frac{20x^2+20xy-y^2}{21x^2-xy}$$

Answer 2

$e^{y/x} \cdot \frac{x-y}{x^2} = 20 - \frac{dy}{dx}$. 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.

Answer 3

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: $$y'=-\frac{F_x}{F_y}=-\frac{e^{y/x}\cdot (-y/x^2)-2}{e^{y/x}\cdot (1/x)+1}=\frac{(2x-y)y+2x^2}{(2x-y)x+x^2}=\frac{2x^2+2xy-y^2}{3x^2-xy}.$$