19. - / 1 points differentiate the function. $p(t)=ln(sqrt{t^{2}+8})$ $p(t)=$
19. - / 1 points differentiate the function. $p(t)=ln(sqrt{t^{2}+8})$ $p(t)=$
Answer
Explanation:
Step1: Simplify the function
Use the property $\ln(\sqrt{x})=\frac{1}{2}\ln(x)$. So $p(t)=\frac{1}{2}\ln(t^{2}+8)$.
Step2: Apply the chain - rule
The derivative of $\ln(u)$ with respect to $t$ is $\frac{u'}{u}$, where $u = t^{2}+8$ and $u'=2t$. The derivative of $\frac{1}{2}\ln(u)$ with respect to $t$ is $\frac{1}{2}\cdot\frac{u'}{u}$. Substitute $u = t^{2}+8$ and $u' = 2t$ into the formula: $p'(t)=\frac{1}{2}\cdot\frac{2t}{t^{2}+8}$.
Step3: Simplify the result
$\frac{1}{2}\cdot\frac{2t}{t^{2}+8}=\frac{t}{t^{2}+8}$.
Answer:
$\frac{t}{t^{2}+8}$