Let $w = \ln ( x y z )$ where x = t, y = t + 1, and z = t + 2. When t = 1, $\frac { d z } { d t }$ is

A) $- \frac { 11 } { 6 }$
B) $\frac { 13 } { 6 }$
C) $- \frac { 13 } { 6 }$
D) $\frac { 17 } { 6 }$
E) $\frac { 11 } { 6 }$

Question

Let $w = \ln ( x y z )$ where x = t, y = t + 1, and z = t + 2. When t = 1, $\frac { d z } { d t }$ is

A) $- \frac { 11 } { 6 }$
B) $\frac { 13 } { 6 }$
C) $- \frac { 13 } { 6 }$
D) $\frac { 17 } { 6 }$
E) $\frac { 11 } { 6 }$

Quizplus · Accepted Answer

Given $w = \ln(xyz)$, with $x = t$, $y = t + 1$, and $z = t + 2$, we differentiate $w$ with respect to $t$ to find $\frac{dw}{dt}$. Using the chain rule and properties of logarithms, we get:$$\frac{dw}{dt} = \frac{1}{xyz} \cdot (yz + xz + xy) \cdot \frac{d}{dt}(t)$$Substituting $x = t$, $y = t + 1$, and $z = t + 2$ and evaluating at $t = 1$:$$\frac{dw}{dt} = \frac{1}{1 \cdot 2 \cdot 3} \cdot ((2 \cdot 3) + (1 \cdot 3) + (1 \cdot 2)) = \frac{1}{6} \cdot (6 + 3 + 2) = \frac{11}{6}$$The question mistakenly asks for $\frac{dz}{dt}$ but provides information to calculate $\frac{dw}{dt}$, leading to the correct answer based on the provided calculation.

Let w=ln⁡(xyz)w = \ln ( x y z )w=ln(xyz) Where X = T, Y = T + 1, and X

Let $w = \ln ( x y z )$ Where X = T, Y = T + 1, and X