# How do you find f'(x) using the definition of a derivative for f(x)=(1-6t)/(5+t)?

$- \frac{31}{5 + t} ^ 2$
$f \left(t\right) = \frac{1 - 6 t}{5 + t}$
$f \left(t + h\right) = \frac{1 + 6 \left(t + h\right)}{5 + \left(t + h\right)}$
$f \left(t + h\right) - f \left(t\right) = \frac{1 - 6 \left(t + h\right)}{5 + \left(t + h\right)} - \frac{1 - 6 t}{5 + t} = \frac{5 + t - 6 \left({t}^{2} + 5 h + 5 t + h t\right) - 5 - t - h + 30 t + 6 {t}^{2} + 6 h t}{\left(5 + t + h\right) \left(5 + t\right)} = \frac{- 31 h}{\left(5 + t + h\right) \left(5 + t\right)}$
$L {t}_{h \to 0} \left(\frac{f \left(t + h\right) - f \left(h\right)}{h}\right) = L {t}_{h \to 0} \frac{- 31 h}{h \left(5 + t + h\right) \left(5 + t\right)} = L {t}_{h \to 0} \frac{- 31}{\left(5 + t + h\right) \left(5 + t\right)} = - \frac{31}{5 + t} ^ 2$