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

Explanation:

Step1: Recall derivative formula

The derivative of $y = \sec^{-1}(u)$ with respect to $t$ is $\frac{dy}{dt}=\frac{1}{|u|\sqrt{u^{2}-1}}\cdot\frac{du}{dt}$ by the chain - rule. Here $u = \frac{1}{4t^{3}}$.

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 into chain - rule formula

Substitute $u = \frac{1}{4t^{3}}$ and $\frac{du}{dt}=-\frac{3}{4t^{4}}$ into $\frac{dy}{dt}=\frac{1}{|u|\sqrt{u^{2}-1}}\cdot\frac{du}{dt}$. We have $\frac{dy}{dt}=\frac{1}{\left|\frac{1}{4t^{3}}\right|\sqrt{\left(\frac{1}{4t^{3}}\right)^{2}-1}}\cdot\left(-\frac{3}{4t^{4}}\right)$. Simplify the expression: [ \begin{align*} \frac{dy}{dt}&=\frac{-3}{4t^{4}\left|\frac{1}{4t^{3}}\right|\sqrt{\left(\frac{1}{4t^{3}}\right)^{2}-1}}\ &=\frac{-3\times4t^{3}}{4t^{4}\sqrt{\left(\frac{1}{4t^{3}}\right)^{2}-1}}\ &=\frac{- 12t^{2}}{\left|\frac{4}{4t^{3}}\right|\sqrt{\left(\frac{1}{4t^{3}}\right)^{2}-1}} \end{align*} ]

Answer:

$\frac{dy}{dt}=\frac{-12t^{2}}{\left|\frac{4}{4t^{3}}\right|\sqrt{\left(\frac{1}{4t^{3}}\right)^{2}-1}}$