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

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})