How do you implicitly differentiate #18=(y-x)^2-ln(xy)#?

Question

Emily Ammerman · Accepted Answer

#(2(y-x)+1/x)/(2(y-x)-1/y )=dy/dx# #18=(y-x)^2-ln(xy)# #0=-2(y-x)+2(y-x)dy/dx - (y/(xy)+x/(xy) dy/dx)# #0=-2(y-x)+2(y-x)dy/dx - 1/x-1/y dy/dx# #2(y-x)+1/x=2(y-x)dy/dx-1/y dy/dx# #2(y-x)+1/x=(2(y-x)-1/y )dy/dx# #(2(y-x)+1/x)/(2(y-x)-1/y )=dy/dx#

Emma Aldridge · Answer

undefined To implicitly differentiate $18=(y-x)^2-\ln(xy)$, follow these steps:

1. Differentiate each term of the equation with respect to $x$.
2. Apply the chain rule when differentiating composite functions.
3. Solve for $\frac{dy}{dx}$ by isolating the terms involving $y$ on one side and the terms involving $x$ on the other side.
4. Express $\frac{dy}{dx}$ in terms of $x$ and $y$.

Differentiating each term:
$$
\begin{align*}
\frac{d}{dx}(18) &= 0 \
\frac{d}{dx}((y-x)^2) &= 2(y-x)\frac{dy}{dx} - 2 \
\frac{d}{dx}(\ln(xy)) &= \frac{1}{xy}(y + x\frac{dy}{dx})
\end{align*}
$$

Substituting into the equation and solving for $\frac{dy}{dx}$:
$$
0 = 2(y-x)\frac{dy}{dx} - 2 - \frac{1}{xy}(y + x\frac{dy}{dx})
$$

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

$$
(2(y-x) - \frac{x}{y})\frac{dy}{dx} = 2 + \frac{y}{xy}
$$

$$
\frac{dy}{dx} = \frac{2 + \frac{y}{xy}}{2(y-x) - \frac{x}{y}}
$$

Cameron Alexander · Answer

undefined To implicitly differentiate $18 = (y - x)^2 - \ln(xy)$ with respect to $x$, you'll differentiate each term with respect to $x$ while treating $y$ as a function of $x$ using the chain rule where necessary. Then, solve for $\frac{dy}{dx}$.

Starting with the given equation:

$$ 18 = (y - x)^2 - \ln(xy) $$

Differentiating each term:

1. Differentiate $(y - x)^2$:
$$ \frac{d}{dx}(y - x)^2 = 2(y - x) \frac{dy}{dx} - 1 $$

2. Differentiate $-\ln(xy)$:
$$ \frac{d}{dx}(-\ln(xy)) = -\frac{1}{xy} \cdot (y + xy') - \frac{1}{xy} \cdot (x \cdot y' + y) $$

Combining terms:

$$ 0 = 2(y - x) \frac{dy}{dx} - 1 - \frac{y}{x} - y' - \frac{x}{y} - x \cdot y' - y $$

Simplify and solve for $y'$:

$$ 0 = 2(y - x) \frac{dy}{dx} - 1 - \frac{y}{x} - y' - \frac{x}{y} - x \cdot y' - y $$

$$ 0 = 2(y - x) \frac{dy}{dx} - y' - x \cdot y' $$

$$ 0 = (2(y - x) - 1 - x) \frac{dy}{dx} - y' $$

$$ \frac{dy}{dx} = \frac{y'}{2(y - x) - 1 - x} $$

That's the implicit differentiation of the given equation with respect to $x$, giving $\frac{dy}{dx}$ in terms of $x$, $y$, and $y'$.