Solve for: g(t)=(3)/(t-4)

Expression: $g\left( t \right)=\frac{ 3 }{ t-4 }$

To find the derivative using the alternative definition, evaluate the limit $\lim_{t \rightarrow a} \left(\frac{ g\left( t \right)-g\left( a \right) }{ t-a }\right)$

$g '\left( a \right)=\lim_{t \rightarrow a} \left(\frac{ \frac{ 3 }{ t-4 }-\frac{ 3 }{ a-4 } }{ t-a }\right)$

Evaluate the limit

$g '\left( a \right)=-\frac{ 3 }{ {a}^{2}-8a+16 }$

The initial variable is $t$, so substitute $t$ for $a$

$g '\left( t \right)=-\frac{ 3 }{ {t}^{2}-8t+16 }$

Use ${a}^{2}-2ab+{b}^{2}={\left( a-b \right)}^{2}$ to factor the expression

$g '\left( t \right)=-\frac{ 3 }{ {\left( t-4 \right)}^{2} }$