Question

use implicit differentiation to find \\( \frac { \partial z } { \partial x } \\) and \\( \frac { \partial z } { \partial y } \\).

\\( e ^ { 2 z } = x y z \\)

Explanation:

Step1: Differentiate both sides with respect to \(x\)

Differentiate \(e^{2z}\) with respect to \(x\) using the chain rule: \(2e^{2z}\frac{\partial z}{\partial x}\).
Differentiate \(xyz\) with respect to \(x\) using the product rule: \(y(z + x\frac{\partial z}{\partial x})\).
So, \(2e^{2z}\frac{\partial z}{\partial x}=y(z + x\frac{\partial z}{\partial x})\).

Step2: Solve for \(\frac{\partial z}{\partial x}\)

Expand the right - hand side: \(2e^{2z}\frac{\partial z}{\partial x}=yz+xy\frac{\partial z}{\partial x}\).
Bring terms with \(\frac{\partial z}{\partial x}\) to one side: \(2e^{2z}\frac{\partial z}{\partial x}-xy\frac{\partial z}{\partial x}=yz\).
Factor out \(\frac{\partial z}{\partial x}\): \(\frac{\partial z}{\partial x}(2e^{2z}-xy)=yz\).
Then \(\frac{\partial z}{\partial x}=\frac{yz}{2e^{2z}-xy}\).

Step3: Differentiate both sides with respect to \(y\)

Differentiate \(e^{2z}\) with respect to \(y\) using the chain rule: \(2e^{2z}\frac{\partial z}{\partial y}\).
Differentiate \(xyz\) with respect to \(y\) using the product rule: \(x(z + y\frac{\partial z}{\partial y})\).
So, \(2e^{2z}\frac{\partial z}{\partial y}=x(z + y\frac{\partial z}{\partial y})\).

Step4: Solve for \(\frac{\partial z}{\partial y}\)

Expand the right - hand side: \(2e^{2z}\frac{\partial z}{\partial y}=xz+xy\frac{\partial z}{\partial y}\).
Bring terms with \(\frac{\partial z}{\partial y}\) to one side: \(2e^{2z}\frac{\partial z}{\partial y}-xy\frac{\partial z}{\partial y}=xz\).
Factor out \(\frac{\partial z}{\partial y}\): \(\frac{\partial z}{\partial y}(2e^{2z}-xy)=xz\).
Then \(\frac{\partial z}{\partial y}=\frac{xz}{2e^{2z}-xy}\).

Answer:

\(\frac{\partial z}{\partial x}=\frac{yz}{2e^{2z}-xy}\), \(\frac{\partial z}{\partial y}=\frac{xz}{2e^{2z}-xy}\)