Questions: Find dy/dt if y=1/(u^2+u) and u=7+4t

Solution Steps

Given: \[ y = \frac{1}{u^2 + u} \]

First, we need to find \(\frac{dy}{du}\). Using the chain rule, we have: \[ \frac{dy}{du} = \frac{d}{du} \left( \frac{1}{u^2 + u} \right) \]

We can use the quotient rule for differentiation, where if \( y = \frac{1}{v} \), then \(\frac{dy}{du} = -\frac{1}{v^2} \frac{dv}{du}\). Here, \( v = u^2 + u \).

\[ v = u^2 + u \] \[ \frac{dv}{du} = 2u + 1 \]

Using the quotient rule: \[ \frac{dy}{du} = -\frac{1}{(u^2 + u)^2} \cdot (2u + 1) \] \[ \frac{dy}{du} = -\frac{2u + 1}{(u^2 + u)^2} \]

Given: \[ u = 7 + 4t \] \[ \frac{du}{dt} = 4 \]

Now, we need to find \(\frac{dy}{dt}\) using the chain rule: \[ \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} \]

Substitute \(\frac{dy}{du}\) and \(\frac{du}{dt}\): \[ \frac{dy}{dt} = -\frac{2u + 1}{(u^2 + u)^2} \cdot 4 \] \[ \frac{dy}{dt} = -\frac{4(2u + 1)}{(u^2 + u)^2} \]

Finally, substitute \( u = 7 + 4t \) into the expression: \[ \frac{dy}{dt} = -\frac{4(2(7 + 4t) + 1)}{((7 + 4t)^2 + (7 + 4t))^2} \] \[ \frac{dy}{dt} = -\frac{4(14 + 8t + 1)}{((7 + 4t)^2 + 7 + 4t)^2} \] \[ \frac{dy}{dt} = -\frac{4(15 + 8t)}{((7 + 4t)^2 + 7 + 4t)^2} \]

Final Answer