Question

if $y=(t^{2}+3t + 2)(3t^{2}+2)$, find $\frac{dy}{dt}$. $\frac{dy}{dt}=square$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $\frac{dy}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}$. Let $u=t^{2}+3t + 2$ and $v = 3t^{2}+2$.

Step2: Find $\frac{du}{dt}$

Differentiate $u=t^{2}+3t + 2$ with respect to $t$. Using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we get $\frac{du}{dt}=2t+3$.

Step3: Find $\frac{dv}{dt}$

Differentiate $v = 3t^{2}+2$ with respect to $t$. Using the power - rule, we get $\frac{dv}{dt}=6t$.

Step4: Substitute into product - rule

$\frac{dy}{dt}=(t^{2}+3t + 2)\times(6t)+(3t^{2}+2)\times(2t + 3)$
$=6t^{3}+18t^{2}+12t+(6t^{3}+9t^{2}+4t + 6)$
$=6t^{3}+18t^{2}+12t+6t^{3}+9t^{2}+4t + 6$
$=12t^{3}+27t^{2}+16t + 6$

Answer:

$12t^{3}+27t^{2}+16t + 6$