The equation of the tangent line to $f ( x ) = 2 \ln x ^ { 3 }$ at x = e is
A) $y = \frac { 6 } { e } x + 12$
B) y = 6x
C) $y = \frac { 6 } { e } x$
D) $y = \frac { 6 } { e } x - 12$

Question

Quizplus · Accepted Answer

To find the equation of the tangent line to $f(x) = 2\ln(x^3)$ at $x = e$, we first need to find the derivative of $f(x)$, which gives us the slope of the tangent line at any point $x$. The derivative of $f(x)$ using the chain rule is $f'(x) = 2 \cdot \frac{1}{x^3} \cdot 3x^2 = \frac{6}{x}$. At $x = e$, the slope is $f'(e) = \frac{6}{e}$. The point on the curve at $x = e$ is $f(e) = 2\ln(e^3) = 2 \cdot 3 = 6$. Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, with $m = \frac{6}{e}$, $x_1 = e$, and $y_1 = 6$, we get $y - 6 = \frac{6}{e}(x - e)$. Simplifying, $y = \frac{6}{e}x + 6 - \frac{6}{e}e = \frac{6}{e}x$, which matches choice C.

The Equation of the Tangent Line To f(x)=2ln⁡x3f ( x ) = 2 \ln x ^ { 3 }f(x)=2lnx3

The Equation of the Tangent Line To $f ( x ) = 2 \ln x ^ { 3 }$