find $\frac{dy}{dt}$. \n$y=(30 + e^{t})ln t$\n$\frac{dy}{dt}=square$\n(type an exact answer.)
find $\frac{dy}{dt}$. \n$y=(30 + e^{t})ln t$\n$\frac{dy}{dt}=square$\n(type an exact answer.)
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, $u = 30 + e^{t}$ and $v=\ln t$.
Step2: Find $\frac{du}{dt}$
Differentiate $u = 30 + e^{t}$ with respect to $t$. Since the derivative of a constant 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: Substitute into product - rule
$\frac{dy}{dt}=(30 + e^{t})\frac{1}{t}+\ln t\cdot e^{t}=\frac{30 + e^{t}}{t}+e^{t}\ln t$.
Answer:
$\frac{30 + e^{t}}{t}+e^{t}\ln t$