Question

determine $\frac{d^{3}y}{dt^{3}}$ if $y = 9^{t}+ln(2.7t)-t^{2}$. attempt 2: 1 attempt remaining.

Explanation:

Step1: Recall power - rule and derivative rules

The power - rule for differentiation is $\frac{d}{dt}(t^n)=nt^{n - 1}$, $\frac{d}{dt}(a^t)=a^t\ln a$ and $\frac{d}{dt}(\ln(bt))=\frac{1}{t}$ for a constant $b$. First, find the first - derivative of $y = 9^t+\ln(2.7t)-t^2$.
$y^\prime=\frac{dy}{dt}=9^t\ln 9+\frac{1}{t}-2t$.

Step2: Find the second - derivative

Differentiate $y^\prime$ with respect to $t$.
$y^{\prime\prime}=\frac{d^2y}{dt^2}=9^t(\ln 9)^2-\frac{1}{t^2}-2$.

Step3: Find the third - derivative

Differentiate $y^{\prime\prime}$ with respect to $t$.
$y^{\prime\prime\prime}=\frac{d^3y}{dt^3}=9^t(\ln 9)^3+\frac{2}{t^3}$.

Answer:

$9^t(\ln 9)^3+\frac{2}{t^3}$