## Calculus: Early Transcendentals 8th Edition

$f''(2)=-\dfrac{4}{27}$
$f(t)=\sqrt{4t+1}$, 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 $f''(2)$: $f''(2)=-\dfrac{4}{\sqrt{[4(2)+1]^{3}}}=-\dfrac{4}{\sqrt{(9)^{3}}}=-\dfrac{4}{9\sqrt{9}}=-\dfrac{4}{(9)(3)}=...$ $...=-\dfrac{4}{27}$