Question

determine $\frac{d^{3}y}{dt^{3}}$ if $y = 8^{t}+ln(3.5t)-t^{2}$. attempt 1: 2 attempts remaining. submit answer next item

Explanation:

Step1: Recall derivative rules

We know that if \(y = a^t\), then \(y^\prime=a^t\ln(a)\), if \(y=\ln(bt)\) then \(y^\prime=\frac{1}{t}\), and if \(y = t^n\) then \(y^\prime=nt^{n - 1}\).

Step2: Find the first - derivative of \(y\)

Given \(y = 8^t+\ln(3.5t)-t^2\).
The derivative of \(8^t\) is \(8^t\ln(8)\), the derivative of \(\ln(3.5t)=\frac{1}{t}\), and the derivative of \(-t^2=-2t\). So \(y^\prime=8^t\ln(8)+\frac{1}{t}-2t\).

Step3: Find the second - derivative of \(y\)

The derivative of \(8^t\ln(8)\) is \(8^t(\ln(8))^2\), the derivative of \(\frac{1}{t}=t^{-1}\) is \(-t^{-2}=-\frac{1}{t^2}\), and the derivative of \(-2t\) is \(-2\). So \(y^{\prime\prime}=8^t(\ln(8))^2-\frac{1}{t^2}-2\).

Step4: Find the third - derivative of \(y\)

The derivative of \(8^t(\ln(8))^2\) is \(8^t(\ln(8))^3\), the derivative of \(-\frac{1}{t^2}=-t^{-2}\) is \(2t^{-3}=\frac{2}{t^3}\), and the derivative of \(-2\) is \(0\). So \(\frac{d^3y}{dt^3}=8^t(\ln(8))^3+\frac{2}{t^3}\).

Answer:

\(8^t(\ln(8))^3+\frac{2}{t^3}\)