Answer
$f''(2)=-\dfrac{4}{27}$
Work Step by Step
$f(t)=\sqrt{4t+1}$, find $f''(2)$
Rewrite $f(t)$ using a rational exponent:
$f(t)=(4t+1)^{1/2}$
Find the first derivative using the chain rule:
$f'(t)=\dfrac{1}{2}(4t+1)^{-1/2}(4)=\dfrac{2}{(4t+1)^{1/2}}$
Apply the quotient rule and the chain rule to evaluate the second derivative:
$f''(t)=\dfrac{(4t+1)^{1/2}(2)'-(2)[(4t+1)^{1/2}]'}{[(4t+1)^{1/2}]^{2}}=...$
$...=\dfrac{(4t+1)^{1/2}(0)-(2)\Big(\dfrac{1}{2}\Big)(4t+1)^{-1/2}(4)}{4t+1}=...$
$...=\dfrac{-4(4t+1)^{-1/2}}{4t+1}=-\dfrac{4}{(4t+1)(4t+1)^{1/2}}=...$
$...=-\dfrac{4}{(4t+1)^{3/2}}=-\dfrac{4}{\sqrt{(4t+1)^{3}}}$
Substitute $t$ by $2$ in $f''(t)$ to obtain $f''(2)$:
$f''(2)=-\dfrac{4}{\sqrt{[4(2)+1]^{3}}}=-\dfrac{4}{\sqrt{(9)^{3}}}=-\dfrac{4}{9\sqrt{9}}=-\dfrac{4}{(9)(3)}=...$
$...=-\dfrac{4}{27}$
