1. Understand the problem:
Given \( y = 2x^{3x} \), we need to find \( \frac{dy}{dx} \) at \( x = 1 \).
2. Use logarithmic differentiation:
First, take the natural logarithm of both sides:
\[ \ln y = \ln(2x^{3x}) = \ln 2 + 3x \ln x \]
Differentiate implicitly with respect to \( x \):
\[ \frac{1}{y} \frac{dy}{dx} = 0 + 3 \ln x + 3x \cdot \frac{1}{x} = 3 \ln x + 3 \]
Multiply both sides by \( y \):
\[ \frac{dy}{dx} = y (3 \ln x + 3) = 2x^{3x} (3 \ln x + 3) \]
3. Evaluate at \( x = 1 \):
\[ \frac{dy}{dx} \bigg|_{x=1} = 2 \cdot 1^{3} (3 \ln 1 + 3) = 2 \cdot 1 (0 + 3) = 6 \]
Correct Answer: (B) 6
Given
$y = 2x^{3x}$, take the natural logarithm of both sides: \[ \ln y = \ln 2 + 3x \ln x \] Differentiate both sides with respect to $x$: \[ \frac{1}{y} \cdot \frac{dy}{dx} = 0 + 3 \ln x + 3 \] Multiply both sides by $y$ to find $\frac{dy}{dx}$: \[ \frac{dy}{dx} = y (3 \ln x + 3) \] Substitute $x = 1$: \[ y = 2(1)^{3(1)} = 2 \] \[ \ln 1 = 0 \] \[ \frac{dy}{dx} = 2 (0 + 3) = 6 \] Hence, the correct answer is 6.