Question

find the derivative of y with respect to t.
y = cot^{-1} \sqrt{t}
\frac{dy}{dt}=\square

Explanation:

Step1: Recall derivative formula

The derivative of $y = \cot^{-1}(u)$ with respect to $u$ is $\frac{dy}{du}=-\frac{1}{1 + u^{2}}$. Here $u = \sqrt{t}$.

Step2: Find derivative of $u$ with respect to $t$

If $u=\sqrt{t}=t^{\frac{1}{2}}$, then $\frac{du}{dt}=\frac{1}{2}t^{-\frac{1}{2}}=\frac{1}{2\sqrt{t}}$ by the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$.

Step3: Use chain - rule

The chain - rule states that $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. Substituting $\frac{dy}{du}=-\frac{1}{1 + u^{2}}$ and $u = \sqrt{t}$ and $\frac{du}{dt}=\frac{1}{2\sqrt{t}}$, we get $\frac{dy}{dt}=-\frac{1}{1+(\sqrt{t})^{2}}\cdot\frac{1}{2\sqrt{t}}$.

Step4: Simplify the expression

Since $(\sqrt{t})^{2}=t$, then $\frac{dy}{dt}=-\frac{1}{2\sqrt{t}(1 + t)}$.

Answer:

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