find.\n$\frac{d^{2}}{dt^{2}}(t^{4}ln t)$\n$\frac{d^{2}}{dt^{2}}(t^{4}ln t)=square$

Explanation:

Step1: Find the first - derivative

Use the product rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = t^{4}$ and $v=\ln t$. The derivative of $u=t^{4}$ is $u^\prime = 4t^{3}$, and the derivative of $v = \ln t$ is $v^\prime=\frac{1}{t}$. Then $\frac{d}{dt}(t^{4}\ln t)=4t^{3}\ln t + t^{4}\cdot\frac{1}{t}=4t^{3}\ln t+t^{3}$.

Step2: Find the second - derivative

Use the sum rule and product rule again. For the first term $y_1 = 4t^{3}\ln t$, by the product rule: if $u = 4t^{3}$ ($u^\prime=12t^{2}$) and $v=\ln t$ ($v^\prime=\frac{1}{t}$), then $y_1^\prime=12t^{2}\ln t+4t^{3}\cdot\frac{1}{t}=12t^{2}\ln t + 4t^{2}$. The derivative of the second term $y_2=t^{3}$ is $y_2^\prime = 3t^{2}$. Then $\frac{d^{2}}{dt^{2}}(t^{4}\ln t)=12t^{2}\ln t+4t^{2}+3t^{2}=12t^{2}\ln t + 7t^{2}$.

Answer:

$12t^{2}\ln t + 7t^{2}$