Question

attempt 1: 10 attempts remaining. find ( g(t) ) given ( g(t) = \frac{5e^t}{t^2 - 19} ). ( g(t) = ) submit answer next item

Explanation:

Step1: Identify the quotient rule

We have \( g(t)=\frac{5e^t}{t^2 - 19} \), so we use the quotient rule for differentiation: if \( g(t)=\frac{u(t)}{v(t)} \), then \( g'(t)=\frac{u'(t)v(t)-u(t)v'(t)}{[v(t)]^2} \). Here, \( u(t) = 5e^t \) and \( v(t)=t^2 - 19 \).

Step2: Find \( u'(t) \) and \( v'(t) \)

The derivative of \( u(t) = 5e^t \) is \( u'(t)=5e^t \) (since the derivative of \( e^t \) is \( e^t \) and the constant multiple rule applies). The derivative of \( v(t)=t^2 - 19 \) is \( v'(t) = 2t \) (using the power rule: derivative of \( t^n \) is \( nt^{n - 1} \), and derivative of a constant is 0).

Step3: Apply the quotient rule

Substitute \( u(t) \), \( u'(t) \), \( v(t) \), and \( v'(t) \) into the quotient rule formula:
\[

$$\begin{align*} g'(t)&=\frac{(5e^t)(t^2 - 19)- (5e^t)(2t)}{(t^2 - 19)^2}\\ &=\frac{5e^t(t^2 - 19 - 2t)}{(t^2 - 19)^2}\\ &=\frac{5e^t(t^2 - 2t - 19)}{(t^2 - 19)^2} \end{align*}$$

\]

Answer:

\( \frac{5e^t(t^2 - 2t - 19)}{(t^2 - 19)^2} \)