We are tasked with finding the derivative $\frac{d}{dt}$ of the equation:

$\frac{xy^3}{1 + y^2} = \frac{8}{5}$

and evaluating $\frac{dy}{dt}$ when $x = 1$, $y = 2$, and $\frac{dx}{dt} = 6$.

Step 1: Implicit Differentiation

We will differentiate both sides of the equation with respect to $t$, using the chain rule where necessary.

The given equation is:

$\frac{xy^3}{1 + y^2} = \frac{8}{5}$

Differentiate both sides with respect to $t$:

$\frac{d}{dt} \left( \frac{xy^3}{1 + y^2} \right) = \frac{d}{dt} \left( \frac{8}{5} \right)$

Since $\frac{8}{5}$ is a constant, its derivative is zero:

$\frac{d}{dt} \left( \frac{xy^3}{1 + y^2} \right) = 0$

Now, we apply the quotient rule to differentiate the left-hand side. Recall the quotient rule:

$\frac{d}{dt} \left( \frac{f(t)}{g(t)} \right) = \frac{f'(t)g(t) - f(t)g'(t)}{(g(t))^2}$

Here, $f(t) = xy^3$ and $g(t) = 1 + y^2$. We differentiate each of these terms with respect to $t$:

Step 1.1: Differentiate $f(t) = xy^3$

We apply the product rule for differentiation:

$\frac{d}{dt} (xy^3) = \frac{d}{dt}(x) \cdot y^3 + x \cdot \frac{d}{dt}(y^3)$

Now, use the chain rule to differentiate $y^3$:

$\frac{d}{dt}(y^3) = 3y^2 \cdot \frac{dy}{dt}$

Thus,

$\frac{d}{dt}(xy^3) = \frac{dx}{dt} \cdot y^3 + x \cdot 3y^2 \cdot \frac{dy}{dt}$

Step 1.2: Differentiate $g(t) = 1 + y^2$

We apply the chain rule:

$\frac{d}{dt}(1 + y^2) = 2y \cdot \frac{dy}{dt}$

Step 1.3: Apply the Quotient Rule

Now we apply the quotient rule:

$\frac{d}{dt} \left( \frac{xy^3}{1 + y^2} \right) = \frac{\left( \frac{dx}{dt} \cdot y^3 + x \cdot 3y^2 \cdot \frac{dy}{dt} \right)(1 + y^2) - xy^3 \cdot 2y \cdot \frac{dy}{dt}}{(1 + y^2)^2}$

Step 2: Evaluate at $x = 1$, $y = 2$, and $\frac{dx}{dt} = 6$

We now substitute the values $x = 1$, $y = 2$, and $\frac{dx}{dt} = 6$ into the equation.

First, substitute $x = 1$, $y = 2$:

• $y^3 = 2^3 = 8$
• $1 + y^2 = 1 + 2^2 = 5$
• $2y = 2 \cdot 2 = 4$
• $3y^2 = 3 \cdot 2^2 = 12$

Now the equation becomes:

$\frac{(6 \cdot 8 + 1 \cdot 12 \cdot \frac{dy}{dt})(5) - 1 \cdot 8 \cdot 4 \cdot \frac{dy}{dt}}{5^2} = 0$

Simplifying:

$\frac{(48 + 12 \frac{dy}{dt})(5) - 32 \frac{dy}{dt}}{25} = 0$

Multiply through by 25:

$(48 + 12 \frac{dy}{dt})(5) - 32 \frac{dy}{dt} = 0$

Simplify further:

$240 + 60 \frac{dy}{dt} - 32 \frac{dy}{dt} = 0$

Combine the terms with $\frac{dy}{dt}$:

$240 + 28 \frac{dy}{dt} = 0$

Solve for $\frac{dy}{dt}$:

$28 \frac{dy}{dt} = -240$

$\frac{dy}{dt} = \frac{-240}{28} = \frac{-60}{7}$

Final Answer:

$\frac{dy}{dt} = -\frac{60}{7}$

Do you have any questions or need further details? Here are some related questions to consider:

1. What would happen if $x$ and $y$ were both functions of $t$?
2. How does implicit differentiation differ from regular differentiation?
3. Can we solve this type of equation if $x$ and $y$ were not given explicitly?
4. How do you handle cases where $y$ appears both inside and outside a function?
5. What happens when the equation involves more than two variables?

Tip: When differentiating implicitly, always be mindful of the chain rule, especially when the variable you are differentiating is inside a function!