Question

find $\frac{dy}{dt}$.
$y=(11 + e^{t})ln t$
$\frac{dy}{dt}=square$ (type an exact answer.)

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $\frac{dy}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}$. Here, $u = 11 + e^{t}$ and $v=\ln t$.

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

Differentiate $u = 11+e^{t}$ with respect to $t$. Since the derivative of a constant ($11$) is $0$ and the derivative of $e^{t}$ with respect to $t$ is $e^{t}$, we have $\frac{du}{dt}=e^{t}$.

Step3: Find $\frac{dv}{dt}$

Differentiate $v = \ln t$ with respect to $t$. The derivative of $\ln t$ with respect to $t$ is $\frac{1}{t}$.

Step4: Calculate $\frac{dy}{dt}$

Using the product - rule $\frac{dy}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}$, substitute $u = 11 + e^{t}$, $\frac{du}{dt}=e^{t}$, $v=\ln t$, and $\frac{dv}{dt}=\frac{1}{t}$:
\[

$$\begin{align*} \frac{dy}{dt}&=(11 + e^{t})\frac{1}{t}+\ln t\cdot e^{t}\\ &=\frac{11 + e^{t}}{t}+e^{t}\ln t \end{align*}$$

\]

Answer:

$\frac{11 + e^{t}}{t}+e^{t}\ln t$