Assume that $x=x(t)$ and $y=y(t)$. Let $y=x^{3}+6$ and $\frac{d x}{d t}=2$ when $x=1$. Find $\frac{d y}{d t}$ when $x=1$

Step 1 ：Given that \(y=x^{3}+6\) and \(\frac{d x}{d t}=2\) when \(x=1\).

Step 2 ：We need to find \(\frac{d y}{d t}\) when \(x=1\).

Step 3 ：First, we can find the derivative of \(y\) with respect to \(x\), denoted as \(\frac{d y}{d x}\).

Step 4 ：By applying the power rule, we get \(\frac{d y}{d x}=3x^{2}\).

Step 5 ：Then, we can use the chain rule to find \(\frac{d y}{d t}\). The chain rule states that \(\frac{d y}{d t}=\frac{d y}{d x} \cdot \frac{d x}{d t}\).

Step 6 ：Substitute \(\frac{d y}{d x}=3x^{2}\) and \(\frac{d x}{d t}=2\) into the equation, we get \(\frac{d y}{d t}=3x^{2} \cdot 2\).

Step 7 ：Finally, substitute \(x=1\) into the equation, we get \(\frac{d y}{d t}=3 \cdot 1^{2} \cdot 2\).

Step 8 ：So, \(\frac{d y}{d t}=6\) when \(x=1\).

Step 9 ：\(\boxed{6}\) is the final answer.