Question

1. -/1 points find $\frac{dy}{dx}$ by implicit differentiation. $x^{2}-18xy + y^{2}=18$ $\frac{dy}{dx}=$ resources read it watch it master it submit answer 2. -/1 points find $\frac{dy}{dx}$ by implicit differentiation. $\frac{x^{2}}{x + y}=y^{2}+9$ $\frac{dy}{dx}=$

Explanation:

Step1: Differentiate each term

Differentiate $x^{2}-18xy + y^{2}=18$ term - by - term with respect to $x$.
The derivative of $x^{2}$ with respect to $x$ is $2x$ (using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$).
For the term $-18xy$, use the product rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$, where $u=-18x$ and $v = y$. So $\frac{d}{dx}(-18xy)=-18y-18x\frac{dy}{dx}$.
The derivative of $y^{2}$ with respect to $x$ is $2y\frac{dy}{dx}$ (using the chain - rule $\frac{d}{dx}(y^{n})=ny^{n - 1}\frac{dy}{dx}$), and the derivative of the constant $18$ with respect to $x$ is $0$.
So we have $2x-18y - 18x\frac{dy}{dx}+2y\frac{dy}{dx}=0$.

Step2: Isolate $\frac{dy}{dx}$

Group the terms with $\frac{dy}{dx}$ on one side:
$2y\frac{dy}{dx}-18x\frac{dy}{dx}=18y - 2x$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(2y - 18x)=18y - 2x$.
Then $\frac{dy}{dx}=\frac{18y - 2x}{2y-18x}=\frac{9y - x}{y - 9x}$.

Step1: Differentiate both sides

First, rewrite the left - hand side as $\frac{x^{2}}{x + y}=x^{2}(x + y)^{-1}$.
Use the product rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$ and the chain rule.
$u = x^{2}$, $\frac{du}{dx}=2x$; $v=(x + y)^{-1}$, $\frac{dv}{dx}=-(x + y)^{-2}(1+\frac{dy}{dx})$.
So $\frac{d}{dx}(\frac{x^{2}}{x + y})=2x(x + y)^{-1}-x^{2}(x + y)^{-2}(1+\frac{dy}{dx})$.
The derivative of the right - hand side $y^{2}+9$ with respect to $x$ is $2y\frac{dy}{dx}$.
We have $\frac{2x}{x + y}-\frac{x^{2}}{(x + y)^{2}}(1+\frac{dy}{dx})=2y\frac{dy}{dx}$.

Step2: Multiply through by $(x + y)^{2}$ to clear the fractions

$2x(x + y)-x^{2}-x^{2}\frac{dy}{dx}=2y(x + y)^{2}\frac{dy}{dx}$.
Expand: $2x^{2}+2xy - x^{2}-x^{2}\frac{dy}{dx}=2y(x^{2}+2xy + y^{2})\frac{dy}{dx}$.
$x^{2}+2xy - x^{2}\frac{dy}{dx}=2x^{2}y\frac{dy}{dx}+4xy^{2}\frac{dy}{dx}+2y^{3}\frac{dy}{dx}$.

Step3: Isolate $\frac{dy}{dx}$

Group the terms with $\frac{dy}{dx}$ on one side:
$x^{2}+2xy=(2x^{2}y + 4xy^{2}+2y^{3}+x^{2})\frac{dy}{dx}$.
Then $\frac{dy}{dx}=\frac{x^{2}+2xy}{x^{2}+2x^{2}y + 4xy^{2}+2y^{3}}$.

Answer:

$\frac{9y - x}{y - 9x}$

For the second problem: