To find the equation of the tangent line to the function $f(x) = \frac{10x}{x^2 - 6}$ at the point $(4, 4)$, we need to follow these steps:

1. Find the derivative $f'(x)$ to determine the slope of the tangent line at $x = 4$.
2. Evaluate $f'(4)$ to get the slope at $x = 4$.
3. Use the point-slope form of the line equation, $y - y_1 = m(x - x_1)$, with $(x_1, y_1) = (4, 4)$ and $m = f'(4)$.

Step 1: Finding the Derivative $f'(x)$

Given: $f(x) = \frac{10x}{x^2 - 6}$ we'll use the quotient rule, which states $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$, where:

• $u = 10x$ and $u' = 10$,
• $v = x^2 - 6$ and $v' = 2x$.

Applying the quotient rule: $f'(x) = \frac{(10)(x^2 - 6) - (10x)(2x)}{(x^2 - 6)^2}$ Simplifying the numerator: $f'(x) = \frac{10x^2 - 60 - 20x^2}{(x^2 - 6)^2} = \frac{-10x^2 - 60}{(x^2 - 6)^2} = \frac{-10(x^2 + 6)}{(x^2 - 6)^2}$

Step 2: Evaluating $f'(4)$

Substitute $x = 4$ into $f'(x)$: $f'(4) = \frac{-10((4)^2 + 6)}{((4)^2 - 6)^2} = \frac{-10(16 + 6)}{(16 - 6)^2} = \frac{-10 \cdot 22}{10^2} = \frac{-220}{100} = -2.2$

So, the slope $m$ at $x = 4$ is $-2.2$.

Step 3: Using the Point-Slope Form

With $m = -2.2$ and point $(4, 4)$, the equation of the tangent line is: $y - 4 = -2.2(x - 4)$ Expanding this: $y = -2.2x + 8.8 + 4 = -2.2x + 12.8$

Final Answer

The equation of the tangent line is: $y = -2.2x + 12.8$

Would you like further details or have any questions?

Here are some related questions for further exploration:

1. How does the quotient rule help in finding derivatives of rational functions?
2. What would happen if we evaluated the derivative at a point where $x^2 = 6$?
3. Why is the point-slope form useful for writing tangent line equations?
4. Can we find the normal line at the point $(4, 4)$ for this function?
5. How does the tangent line differ from the secant line through two points on the curve?

Tip: When finding tangent lines, make sure to correctly apply the derivative rules to obtain the slope at the specified point.