QuestionFind $\frac{d x}{d t}, \frac{d y}{d t}$, and $\frac{d y}{d x}$ $$\begin{array}{l} \frac{d x}{d x}=8 t e^{t}, \quad y=2 t+\sin (t) \\ \frac{d y}{d t}=\square \\ \frac{d y}{d x}=\square \end{array}$$

Studdy Solution
Use the chain rule to find $\frac{d y}{d x}$.
The chain rule states that:
$\frac{d y}{d x} = \frac{d y}{d t} \cdot \frac{d t}{d x}$
Since we have $\frac{d x}{d t}$, we can find $\frac{d t}{d x}$ as the reciprocal of $\frac{d x}{d t}$:
$\frac{d t}{d x} = \frac{1}{\frac{d x}{d t}} = \frac{1}{8e^t + 8t e^t}$
Now substitute into the chain rule:
$\frac{d y}{d x} = \left(2 + \cos(t)\right) \cdot \frac{1}{8e^t + 8t e^t}$
The derivatives are:
$\frac{d x}{d t} = 8e^t + 8t e^t$
$\frac{d y}{d t} = 2 + \cos(t)$
$\frac{d y}{d x} = \frac{2 + \cos(t)}{8e^t + 8t e^t}$