question 13 (6.25 points)\nfind $\frac{partial z}{partial x}$.\n$9x^{2}+8y^{2}-3z^{2}=2x(y + z)$

Explanation:

Step1: Differentiate both sides with respect to x

Differentiate $9x^{2}+8y^{2}-3z^{2}=2x(y + z)$ term - by - term. The derivative of $9x^{2}$ with respect to $x$ is $18x$ (using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$). Since $y$ is treated as a constant, the derivative of $8y^{2}$ with respect to $x$ is $0$. For $-3z^{2}$, using the chain - rule $\frac{d}{dx}(u^{2}) = 2u\frac{du}{dx}$, we get $-6z\frac{\partial z}{\partial x}$. The right - hand side: Using the product rule $(uv)^\prime=u^\prime v+uv^\prime$ where $u = 2x$ and $v=y + z$, we have $\frac{d}{dx}(2x(y + z))=2(y + z)+2x(\frac{\partial z}{\partial x})$. So, $18x-6z\frac{\partial z}{\partial x}=2(y + z)+2x\frac{\partial z}{\partial x}$.

Step2: Isolate $\frac{\partial z}{\partial x}$

Move all terms with $\frac{\partial z}{\partial x}$ to one side: $-6z\frac{\partial z}{\partial x}-2x\frac{\partial z}{\partial x}=2(y + z)-18x$. Factor out $\frac{\partial z}{\partial x}$: $\frac{\partial z}{\partial x}(-6z - 2x)=2y+2z - 18x$. Then $\frac{\partial z}{\partial x}=\frac{2y + 2z-18x}{-6z - 2x}=\frac{9x - y - z}{3z + x}$.

Answer:

$\frac{9x - y - z}{3z + x}$