Question

find the derivative of y with respect to t.
y = \csc^{-1}(e^{10t})
\frac{dy}{dt}=\square

Explanation:

Step1: Recall derivative formula

The derivative of $y = \csc^{-1}(u)$ with respect to $u$ is $\frac{dy}{du}=-\frac{1}{|u|\sqrt{u^{2}-1}}$, and by the chain - rule, if $y=\csc^{-1}(u)$ and $u = e^{10t}$, then $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$.

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

If $u = e^{10t}$, then by the rule that the derivative of $e^{at}$ with respect to $t$ is $ae^{at}$, we have $\frac{du}{dt}=10e^{10t}$.

Step3: Apply the chain - rule

Substitute $u = e^{10t}$ and $\frac{du}{dt}=10e^{10t}$ into the chain - rule formula. Since $u = e^{10t}>0$ for all real $t$, we have:
\[

$$\begin{align*} \frac{dy}{dt}&=-\frac{1}{|e^{10t}|\sqrt{(e^{10t})^{2}-1}}\cdot10e^{10t}\\ &=-\frac{10e^{10t}}{e^{10t}\sqrt{e^{20t}-1}}\\ &=-\frac{10}{\sqrt{e^{20t}-1}} \end{align*}$$

\]

Answer:

$-\frac{10}{\sqrt{e^{20t}-1}}$