To find the equation of the tangent line to the graph of $\frac{4x}{2y^3 - 5y^2} = -4$ at the point $(7, -1)$, follow these steps:

Step 1: Implicit Differentiation

We need to differentiate both sides of the equation implicitly with respect to $x$.

The equation is:

$\frac{4x}{2y^3 - 5y^2} = -4$

Differentiate both sides with respect to $x$, applying the quotient rule on the left side.

Let: $f(x, y) = \frac{4x}{2y^3 - 5y^2}$

Use the quotient rule for differentiation: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$ where $u = 4x$ and $v = 2y^3 - 5y^2$.

First, differentiate $u = 4x$: $u' = 4$

Now, differentiate $v = 2y^3 - 5y^2$ implicitly using the chain rule: $v' = 6y^2 \frac{dy}{dx} - 10y \frac{dy}{dx}$

Now apply the quotient rule: $\frac{d}{dx} \left( \frac{4x}{2y^3 - 5y^2} \right) = \frac{(4)(2y^3 - 5y^2) - (4x)(6y^2 \frac{dy}{dx} - 10y \frac{dy}{dx})}{(2y^3 - 5y^2)^2}$

Set this equal to 0 because the derivative of the constant -4 is zero:

$\frac{(4)(2y^3 - 5y^2) - (4x)(6y^2 \frac{dy}{dx} - 10y \frac{dy}{dx})}{(2y^3 - 5y^2)^2} = 0$

Step 2: Plug in the Point (7, -1)

Now substitute $x = 7$ and $y = -1$ into the equation to find $\frac{dy}{dx}$, which is the slope of the tangent line.

Evaluate $2y^3 - 5y^2$ at $y = -1$: $2(-1)^3 - 5(-1)^2 = -2 - 5 = -7$

Substitute these values into the equation:

$\frac{(4)(-7) - (4 \times 7)(6(-1)^2 \frac{dy}{dx} - 10(-1) \frac{dy}{dx})}{(-7)^2} = 0$

Simplify: $\frac{-28 - 28(6 \frac{dy}{dx} + 10 \frac{dy}{dx})}{49} = 0$

$-28 - 28(16 \frac{dy}{dx}) = 0$

$-28 = 448 \frac{dy}{dx}$

Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-28}{448} = -\frac{1}{16}$

So, the slope of the tangent line at $(7, -1)$ is $m = -\frac{1}{16}$.

Step 3: Equation of the Tangent Line

Now that we have the slope $m = -\frac{1}{16}$, we can use the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$ where $(x_1, y_1) = (7, -1)$ and $m = -\frac{1}{16}$.

Substitute the values: $y + 1 = -\frac{1}{16}(x - 7)$

Simplify: $y + 1 = -\frac{1}{16}x + \frac{7}{16}$

$y = -\frac{1}{16}x + \frac{7}{16} - 1$

$y = -\frac{1}{16}x - \frac{9}{16}$

Final Answer:

The equation of the tangent line is:

$y = -\frac{1}{16}x - \frac{9}{16}$

Would you like more details or have any further questions?

Here are 5 related questions you might find interesting:

1. How do you find the normal line to a curve at a given point?
2. What are the steps to solve an implicit differentiation problem?
3. How does the quotient rule work in differentiation?
4. How do you apply the chain rule in implicit differentiation problems?
5. How would the process change if the function were explicit rather than implicit?

Tip: Implicit differentiation is particularly useful for finding slopes of curves that cannot easily be solved for $y$ as a function of $x$.