Question

hw 12 - higher order derivatives section 2.7: probl (1 point) suppose (f(t)=sqrt{13t^{2}+15}). (f(t)=square) (f(t)=square)

Explanation:

Step1: Rewrite the function

Rewrite $f(t)=\sqrt{13t^{2}+15}=(13t^{2}+15)^{\frac{1}{2}}$.

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

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u = 13t^{2}+15$, then $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$.
$\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dt}=26t$.
So, $f^{\prime}(t)=\frac{1}{2}(13t^{2}+15)^{-\frac{1}{2}}\cdot26t=\frac{13t}{\sqrt{13t^{2}+15}}$.

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 $u = 13t$, $u^{\prime}=13$, $v=\sqrt{13t^{2}+15}=(13t^{2}+15)^{\frac{1}{2}}$, and $v^{\prime}=\frac{13t}{\sqrt{13t^{2}+15}}$ (from Step 2).
\[

$$\begin{align*} f^{\prime\prime}(t)&=\frac{13\sqrt{13t^{2}+15}-13t\cdot\frac{13t}{\sqrt{13t^{2}+15}}}{13t^{2}+15}\\ &=\frac{13(13t^{2}+15)-169t^{2}}{(13t^{2}+15)^{\frac{3}{2}}}\\ &=\frac{169t^{2}+195 - 169t^{2}}{(13t^{2}+15)^{\frac{3}{2}}}\\ &=\frac{195}{(13t^{2}+15)^{\frac{3}{2}}} \end{align*}$$

\]

Answer:

$f^{\prime}(t)=\frac{13t}{\sqrt{13t^{2}+15}}$, $f^{\prime\prime}(t)=\frac{195}{(13t^{2}+15)^{\frac{3}{2}}}$