Question

attempt 1: 10 attempts remaining. find $g(t)$ given $g(t) = \frac{19e^t}{t^3 - 5}$. $g(t) = $ submit answer next item

Explanation:

Step1: Identify the quotient rule

The function \( g(t) = \frac{19e^t}{t^3 - 5} \) is a quotient of two functions, \( u(t) = 19e^t \) and \( v(t) = t^3 - 5 \). The quotient rule states that if \( g(t) = \frac{u(t)}{v(t)} \), then \( g'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2} \).

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

• For \( u(t) = 19e^t \), the derivative \( u'(t) = 19e^t \) (since the derivative of \( e^t \) is \( e^t \) and the constant multiple rule applies).
• For \( v(t) = t^3 - 5 \), the derivative \( v'(t) = 3t^2 \) (using the power rule: derivative of \( t^n \) is \( nt^{n - 1} \), and the 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:
\[
g'(t) = \frac{(19e^t)(t^3 - 5) - (19e^t)(3t^2)}{(t^3 - 5)^2}
\]
Factor out \( 19e^t \) from the numerator:
\[
g'(t) = \frac{19e^t(t^3 - 5 - 3t^2)}{(t^3 - 5)^2}
\]
Simplify the numerator:
\[
g'(t) = \frac{19e^t(t^3 - 3t^2 - 5)}{(t^3 - 5)^2}
\]

Answer:

\( \frac{19e^t(t^3 - 3t^2 - 5)}{(t^3 - 5)^2} \)