find the derivative of y with respect to t.\ny = \\sec^{-1}(\\frac{1}{4t^{3}}), 0 < t < \\sqrt3{\\frac{1}{4}}…

Explanation:

Step1: Recall derivative formula

The derivative of $y = \sec^{-1}(u)$ with respect to $x$ is $\frac{dy}{dx}=\frac{1}{|u|\sqrt{u^{2}-1}}\cdot\frac{du}{dx}$, where $|u|> 1$. Let $u = \frac{1}{4t^{3}}$.

Step2: Find $\frac{du}{dt}$

Using the power - rule for differentiation, if $u=\frac{1}{4t^{3}}=\frac{1}{4}t^{-3}$, then $\frac{du}{dt}=\frac{1}{4}\times(- 3)t^{-4}=-\frac{3}{4t^{4}}$.

Step3: Substitute $u$ and $\frac{du}{dt}$ into the formula

We have $|u|=\left|\frac{1}{4t^{3}}\right|$, and since $0 < t<\sqrt[3]{\frac{1}{4}}$, then $u=\frac{1}{4t^{3}}>1$. [ \begin{align*} \frac{dy}{dt}&=\frac{1}{\left|\frac{1}{4t^{3}}\right|\sqrt{\left(\frac{1}{4t^{3}}\right)^{2}-1}}\cdot\left(-\frac{3}{4t^{4}}\right)\ &=\frac{4t^{3}}{\sqrt{\frac{1}{16t^{6}} - 1}}\cdot\left(-\frac{3}{4t^{4}}\right)\ &=-\frac{3}{t\sqrt{\frac{1 - 16t^{6}}{16t^{6}}}}\ &=-\frac{12t^{2}}{\sqrt{1 - 16t^{6}}} \end{align*} ]

Answer:

$-\frac{12t^{2}}{\sqrt{1 - 16t^{6}}}$