Question

If $y = x \tan y$, then $\frac{dy}{dx} =$

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

Hint:

When differentiating an implicit equation that can easily be rearranged to isolate $x$ as $x = \frac{y}{\tan y} = y \cot y$, it can be much faster to find $\frac{dx}{dy}$ first using the standard quotient or product rule, and then take the reciprocal to get $\frac{dy}{dx} = \frac{1}{dx/dy}$!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the first derivative $\frac{dy}{dx}$ of an implicitly defined trigonometric function $y = x \tan y$.

Step 2: Key Formula or Approach:
We will differentiate both sides of the implicit equation with respect to $x$ using the product rule on the right-hand side, then isolate all terms containing $\frac{dy}{dx}$ on one side to solve for it.

Step 3: Detailed Explanation:
Given the equation:
$$y = x \tan y$$ Differentiating both sides with respect to $x$:
$$\frac{dy}{dx} = \frac{d}{dx}(x \cdot \tan y)$$ Applying the product rule ($\frac{d}{dx}[uv] = u\frac{dv}{dx} + v\frac{du}{dx}$):
$$\frac{dy}{dx} = x \cdot \left(\sec^2 y \cdot \frac{dy}{dx}\right) + \tan y \cdot (1)$$ $$\frac{dy}{dx} = x \sec^2 y \frac{dy}{dx} + \tan y$$ Group all terms containing $\frac{dy}{dx}$ on the left-hand side:
$$\frac{dy}{dx} - x \sec^2 y \frac{dy}{dx} = \tan y$$ $$\frac{dy}{dx}(1 - x \sec^2 y) = \tan y$$ $$\frac{dy}{dx} = \frac{\tan y}{1 - x \sec^2 y}$$ To match the choices provided in the options, let's eliminate the trigonometric terms using our original equation relationships ($\tan y = \frac{y}{x}$ and $\sec^2 y = 1 + \tan^2 y = 1 + \frac{y^2}{x^2}$):
$$\frac{dy}{dx} = \frac{\frac{y}{x}}{1 - x\left(1 + \frac{y^2}{x^2}\right)} = \frac{\frac{y}{x}}{1 - x - \frac{y^2}{x}}$$ Multiply the numerator and the denominator by $x$ to clear the fractions:
$$\frac{dy}{dx} = \frac{y}{x - x^2 - y^2}$$ This matches option (B).

Step 4: Final Answer:
The derivative $\frac{dy}{dx}$ is equal to $\frac{y}{x - x^2 - y^2}$, which corresponds to option (B).