Question

find $\frac{dy}{dt}$. $y = 3t(2t^{2}-5)^{4}$ $\frac{dy}{dt}=\square$

Explanation:

Step1: Apply the product rule

The product rule states that if \(y = uv\), then \(\frac{dy}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}\). Let \(u = 3t\) and \(v=(2t^{2}-5)^{4}\).
First, find \(\frac{du}{dt}\):
\(\frac{du}{dt}=\frac{d}{dt}(3t)=3\)
Next, find \(\frac{dv}{dt}\) using the chain rule. Let \(w = 2t^{2}-5\), so \(v = w^{4}\).
The chain rule: \(\frac{dv}{dt}=\frac{dv}{dw}\cdot\frac{dw}{dt}\)
\(\frac{dv}{dw}=4w^{3}\) and \(\frac{dw}{dt}=4t\)
So \(\frac{dv}{dt}=4(2t^{2}-5)^{3}\cdot4t = 16t(2t^{2}-5)^{3}\)

Step2: Substitute into the product rule formula

\(\frac{dy}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}\)
\(\frac{dy}{dt}=3t\cdot16t(2t^{2}-5)^{3}+(2t^{2}-5)^{4}\cdot3\)
\(=48t^{2}(2t^{2}-5)^{3}+3(2t^{2}-5)^{4}\)
Factor out \(3(2t^{2}-5)^{3}\):
\(=3(2t^{2}-5)^{3}[16t^{2}+(2t^{2}-5)]\)
\(=3(2t^{2}-5)^{3}(18t^{2}-5)\)

Answer:

\(3(2t^{2}-5)^{3}(18t^{2}-5)\)