Question

find (g(t)) given (g(t)=\frac{17e^{t}}{t^{2}-7}). (g(t)=)

Explanation:

Step1: Identify quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 17e^{t}$ and $v=t^{2}-7$.

Step2: Differentiate $u$ and $v$

The derivative of $u = 17e^{t}$ with respect to $t$ is $u'=17e^{t}$ (since the derivative of $e^{t}$ is $e^{t}$ and the constant multiple rule). The derivative of $v=t^{2}-7$ with respect to $t$ is $v' = 2t$.

Step3: Apply the quotient - rule

$g'(t)=\frac{(17e^{t})(t^{2}-7)-(17e^{t})(2t)}{(t^{2}-7)^{2}}=\frac{17e^{t}(t^{2}-2t - 7)}{(t^{2}-7)^{2}}$.

Answer:

$\frac{17e^{t}(t^{2}-2t - 7)}{(t^{2}-7)^{2}}$