Question

For the equation $yz - \ln(z) = x + y$, where $z$ is a function of two independent variables $x$ and $y$ and the partial derivatives exist. The value of $\frac{\partial z}{\partial x}$ is given by}

• A. $\frac{x}{yz - 1}$
• B. $\frac{y}{yz - 1}$
• C. $\frac{1}{yz - 1}$
• D. $\frac{z}{yz - 1}$

Hint:

Always keep track of signs when applying the implicit differentiation formula:
\[ \frac{\partial z}{\partial x} = -\frac{F_x}{F_z} \]

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We can find the partial derivative of an implicitly defined function of several variables by using the Implicit Function Theorem.
Key Formula or Approach:
For a function $F(x, y, z) = 0$, the partial derivative $\frac{\partial z}{\partial x}$ is given by:
\[ \frac{\partial z}{\partial x} = -\frac{\partial F / \partial x}{\partial F / \partial z} \]

Step 2: Detailed Explanation:
Let us write the given equation as a function $F(x, y, z) = 0$:
\[ F(x, y, z) = yz - \ln z - x - y = 0 \]
Now, calculate the partial derivative of $F$ with respect to $x$ (treating $y$ and $z$ as constants):
\[ \frac{\partial F}{\partial x} = -1 \]
Next, calculate the partial derivative of $F$ with respect to $z$ (treating $x$ and $y$ as constants):
\[ \frac{\partial F}{\partial z} = y - \frac{1}{z} = \frac{yz - 1}{z} \]
Substitute these two derivatives into the implicit derivative formula:
\[ \frac{\partial z}{\partial x} = -\frac{-1}{\left(\frac{yz - 1}{z}\right)} = \frac{z}{yz - 1} \]
Therefore, the value of $\frac{\partial z}{\partial x}$ is $\frac{z}{yz - 1}$.

Step 3: Final Answer
The correct option is (D).