Question

find the first and second derivatives. s = 4t^3 - 3t^4
\frac{ds}{dt}=\square

Explanation:

Step1: Apply power - rule for first derivative

The power - rule states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$. For $s = 4t^{3}-3t^{4}$, the first derivative $\frac{ds}{dt}$ is:
$\frac{ds}{dt}=4\times3t^{3 - 1}-3\times4t^{4 - 1}=12t^{2}-12t^{3}$

Step2: Apply power - rule for second derivative

Differentiate $\frac{ds}{dt}=12t^{2}-12t^{3}$ with respect to $t$. Using the power - rule again, we get $\frac{d^{2}s}{dt^{2}}=12\times2t^{2 - 1}-12\times3t^{3 - 1}=24t - 36t^{2}$

Answer:

First derivative $\frac{ds}{dt}=12t^{2}-12t^{3}$, Second derivative $\frac{d^{2}s}{dt^{2}}=24t - 36t^{2}$