Question

hw 12 - higher order derivatives section 2.7: problem 8 (1 point) suppose (f(t)=sqrt{8t^{2}+3}). (f(t)=square) (f(t)=square) note: you can earn partial credit on this problem. preview my answers submit answers you have attempted this problem 0 times. you have 6 attempts remaining.

Explanation:

Step1: Rewrite the function

Rewrite $f(t)=\sqrt{8t^{2}+3}=(8t^{2}+3)^{\frac{1}{2}}$.

Step2: Find the first - derivative using the chain - rule

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u = 8t^{2}+3$, then $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$.
$\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dt}=16t$.
So $f^{\prime}(t)=\frac{1}{2}(8t^{2}+3)^{-\frac{1}{2}}\cdot16t=\frac{8t}{\sqrt{8t^{2}+3}}$.

Step3: Find the second - derivative using the quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^{\prime}=\frac{u^{\prime}v - uv^{\prime}}{v^{2}}$, where $u = 8t$, $u^{\prime}=8$, $v=\sqrt{8t^{2}+3}=(8t^{2}+3)^{\frac{1}{2}}$, and $v^{\prime}=\frac{8t}{\sqrt{8t^{2}+3}}$ (from the chain - rule as above).
$f^{\prime\prime}(t)=\frac{8\sqrt{8t^{2}+3}-8t\cdot\frac{8t}{\sqrt{8t^{2}+3}}}{8t^{2}+3}=\frac{8(8t^{2}+3)-64t^{2}}{(8t^{2}+3)^{\frac{3}{2}}}=\frac{24}{(8t^{2}+3)^{\frac{3}{2}}}$.

Answer:

$f^{\prime}(t)=\frac{8t}{\sqrt{8t^{2}+3}}$
$f^{\prime\prime}(t)=\frac{24}{(8t^{2}+3)^{\frac{3}{2}}}$