#### Answer

$f'(x)=5t+6$

#### Work Step by Step

$f(t)=2.5t^{2}+6t$
Derivative using the definition: $f'(t)=\lim\limits_{h \to 0}\dfrac{f(t+h)-f(t)}{h}$
To find $f(t+h)$, wherever you see $t$ in the function, plug in $t+h$
$f(t+h)=2.5(t+h)^{2}+6(t+h)=2.5(t^{2}+2th+h^{2})+6t+6h=...$
$...=2.5t^{2}+5th+2.5h^{2}+6t+6h$
Let's plug in the components of the formula:
$f'(t)=\lim\limits_{h \to 0}\dfrac{f(t+h)-f(x)}{h}=\lim\limits_{h \to 0}\dfrac{2.5t^{2}+5th+2.5h^{2}+6t+6h-2.5t^{2}-6t}{h}=\lim\limits_{h \to 0}\dfrac{5th+2.5h^{2}+6h}{h}$
Take out common factor $h$
$\lim\limits_{h \to 0}\dfrac{h(5t+2.5h+6)}{h}=\lim\limits_{h \to 0}5t+2.5h+6$
Apply direct substitution:
$5t+2.5(0)+6=5t+6$
$f'(t)=5t+6$