Find the second order derivative of y=9 log⁡ t³

The correct option is –(27/t²).

Step-by-step Derivation:

We are given the function:
$y=9\log \left(t^3\right)$
Here, the logarithm is the natural logarithm.

Step 1: Simplify the expression using logarithmic properties
Using the power rule of logarithms, $\log \left(a^b\right)=b\log \left(a\right)$, we can rewrite the equation as:
$y=9\times 3\log \left(t\right)$
$y=27\log \left(t\right)$

Step 2: Find the first-order derivative
Differentiating both sides with respect to t, we get:
$\frac{dy}{dt}=\frac{d}{dt}[27\log (t\left)\right]$
Since the derivative of $\log \left(t\right)$ with respect to t is $\frac{1}{t}$:
$\frac{dy}{dt}=\frac{27}{t}=27t^{-1}$

Step 3: Find the second-order derivative
Differentiating $\frac{dy}{dt}$ once more with respect to t:
$\frac{d^{2}y}{d{t}^2}=\frac{d}{dt}\left(27t^{-1}\right)$
Using the power rule of differentiation, $\frac{d}{dt}\left(t^n\right)=n{t}^{n-1}$, we substitute $n=-1$:
$\frac{d^{2}y}{d{t}^2}=27\times (-1)t^{-2}$
$\frac{d^{2}y}{d{t}^2}=-\frac{27}{t^2}$

Therefore, the second-order derivative is indeed –(27/t²).