Answer
$f'(t) = -\dfrac{5(2t+3)}{2\sqrt{ t^2+3t+1}}=-\dfrac{10t+15}{2\sqrt{ t^2+3t+1}}$
Work Step by Step
In order to derivate this function you have to apply the chain rule
Let's make an «u» substitution to make it easier
$u = t^2+3t+1$
$f(u) = u^{-5/2}$
Derivate the function:
$f'(u) = -\dfrac{5}{2} u^{7/2}u'$
Now let's find u'
$u' = 2t+3$
Then undo the substitution, simplify and get the answer:
$f'(t) = -\dfrac{5(2t+3)}{2\sqrt{ t^2+3t+1}}=-\dfrac{10t+15}{2\sqrt{ t^2+3t+1}}$
