Question

If $e^{x + y} = 3x$, then $\frac{dy}{dx}$ is equal to:

• A. $\frac{3}{x} - 1$
• B. $\frac{1}{x} - 1$
• C. $\frac{1 - e^{x+y}}{e^{x+y}}$
• D. $\frac{3 - e^x}{e^y}$

Hint:

Taking logs first completely avoids having to deal with chain-rule terms like $e^{x+y}\cdot(1 + \frac{dy}{dx})$, making the algebra clean and fast!

Answer 1

The Correct Option is B

Concept: To find $\frac{dy}{dx}$ from an implicit equation where a variable is in the exponent, we can either use implicit differentiation directly or take the natural logarithm on both sides first to simplify the expression. Taking logs is usually much faster and less prone to errors.

Step 1: Take the natural logarithm on both sides.
The given equation is: \[ e^{x + y} = 3x \] Taking the natural log ($\ln$) of both sides: \[ \ln\left(e^{x + y}\right) = \ln(3x) \] Since $\ln(e^u) = u$, the left side simplifies directly to: \[ x + y = \ln(3x) \] Using logarithmic expansion rules, $\ln(3x) = \ln 3 + \ln x$: \[ x + y = \ln 3 + \ln x \]

Step 2: Differentiate implicitly with respect to $x$.
Now, differentiate each term with respect to $x$: \[ \frac{d}{dx}(x) + \frac{d}{dx}(y) = \frac{d}{dx}(\ln 3) + \frac{d}{dx}(\ln x) \] Evaluating these derivatives: - $\frac{d}{dx}(x) = 1$ - $\frac{d}{dx}(y) = \frac{dy}{dx}$ - $\frac{d}{dx}(\ln 3) = 0$ (since $\ln 3$ is a constant) - $\frac{d}{dx}(\ln x) = \frac{1}{x}$ Substituting these values back gives: \[ 1 + \frac{dy}{dx} = 0 + \frac{1}{x} \]

Step 3: Isolate $\frac{dy{dx}$.}
Subtracting 1 from both sides to solve for $\frac{dy}{dx}$: \[ \frac{dy}{dx} = \frac{1}{x} - 1 \] This matches option (B).