Question

If the equation $f(x, y) = C$ defines $y$ implicitly as a differentiable function of $x$, and if $\frac{\partial f}{\partial y} \neq 0$, then $\frac{dy}{dx}$ is equal to}

• A. $\frac{\partial f / \partial x}{\partial f / \partial y}$
• B. $-\frac{\partial f / \partial x}{\partial f / \partial y}$
• C. $\frac{\partial^2 f}{\partial x^2}$
• D. $\frac{\partial^2 f / \partial^2 f}{\partial x^2 / \partial y^2}$

Hint:

Always remember the negative sign in the implicit differentiation formula:
\[ \frac{dy}{dx} = -\frac{f_x}{f_y} \]

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The Implicit Function Theorem provides a formula to calculate the total derivative $\frac{dy}{dx}$ of an implicitly defined function.

Step 2: Detailed Explanation:
Let the function be defined by:
\[ f(x, y) = C \]
Differentiate both sides of the equation with respect to $x$ using the total derivative chain rule:
\[ \frac{\partial f}{\partial x} \frac{dx}{dx} + \frac{\partial f}{\partial y} \frac{dy}{dx} = 0 \]
Since $\frac{dx}{dx} = 1$:
\[ \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx} = 0 \]
Isolate the term containing $\frac{dy}{dx}$:
\[ \frac{\partial f}{\partial y} \frac{dy}{dx} = -\frac{\partial f}{\partial x} \]
Divide by $\frac{\partial f}{\partial y}$ (given that $\frac{\partial f}{\partial y} \neq 0$):
\[ \frac{dy}{dx} = -\frac{\partial f / \partial x}{\partial f / \partial y} \]
Therefore, the derivative is equal to $-\frac{\partial f / \partial x}{\partial f / \partial y}$.

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