find the derivative of y with respect to t. y = arccot √(3t) dy/dt = -3/(1 + 3t(2√(3t)))

Explanation:

Step1: Recall derivative of arccot(u)

The derivative of $y = \text{arccot}(u)$ with respect to $u$ is $y^\prime=-\frac{1}{1 + u^{2}}$. Here $u = \sqrt{3t}$.

Step2: Apply chain - rule

The chain - rule states that $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. First, find $\frac{du}{dt}$. Since $u=\sqrt{3t}=(3t)^{\frac{1}{2}}$, then $\frac{du}{dt}=\frac{3}{2\sqrt{3t}}$ by the power rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$. Second, $\frac{dy}{du}=-\frac{1}{1 + u^{2}}=-\frac{1}{1+3t}$. Then $\frac{dy}{dt}=-\frac{1}{1 + 3t}\cdot\frac{3}{2\sqrt{3t}}=-\frac{3}{2\sqrt{3t}(1 + 3t)}$.

Answer:

$-\frac{3}{2\sqrt{3t}(1 + 3t)}$