$\text { Find } \frac { d y } { d x } \text {. }$
$x = \sqrt [ 12 ] { t }$
$y = 6 - t$
A)
$\frac { d y } { d x } = - 12 t ^ { \frac { 13 } { 12 } }$
B)
$\frac { d y } { d x } = - 12 t ^ { \frac { 11 } { 12 } }$
C)
$\frac { d y } { d x } = - 6 t ^ { \frac { 11 } { 12 } }$
D)
$\frac { d y } { d x } = 6 t ^ { \frac { 13 } { 12 } }$
E)
$\frac { d y } { d x } = 12 t ^ { \frac { 11 } { 12 } }$

Question

Quizplus · Accepted Answer

To find $\frac{dy}{dx}$, we use the chain rule, which states $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$. First, differentiate $y$ with respect to $t$, which gives $\frac{dy}{dt} = -1$. Then, differentiate $x$ with respect to $t$, $x = t^{\frac{1}{12}}$, which gives $\frac{dx}{dt} = \frac{1}{12}t^{-\frac{11}{12}}$. To find $\frac{dt}{dx}$, we take the reciprocal of $\frac{dx}{dt}$, which is $12t^{\frac{11}{12}}$. Therefore, $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = -1 \cdot 12t^{\frac{11}{12}} = -12t^{\frac{11}{12}}$.

Find dydx. \text { Find } \frac { d y } { d x } \text {. } Find dxdy​.

$\text { Find } \frac { d y } { d x } \text {. }$