Given the function $f(x) = \frac{x}{x-2}$, we want to compute the third derivative, $\frac{d^3 y}{dx^3}$, of the function.

Step 1: First Derivative

We can use the quotient rule for the first derivative of $f(x) = \frac{u(x)}{v(x)}$, where $u(x) = x$ and $v(x) = x-2$. The quotient rule states:

$f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}$

Applying this to our function:

• $u(x) = x$, so $u'(x) = 1$
• $v(x) = x-2$, so $v'(x) = 1$

$f'(x) = \frac{(x-2)(1) - (x)(1)}{(x-2)^2} = \frac{x-2 - x}{(x-2)^2} = \frac{-2}{(x-2)^2}$

Step 2: Second Derivative

Now, we differentiate the first derivative $f'(x) = \frac{-2}{(x-2)^2}$. We apply the chain rule here. Let $g(x) = (x-2)^2$, so the derivative of $g(x)$ is:

$g'(x) = 2(x-2)$

Thus, using the chain rule:

$f''(x) = \frac{d}{dx} \left( \frac{-2}{(x-2)^2} \right) = -2 \cdot \frac{-2 \cdot 2(x-2)}{(x-2)^4} = \frac{4(x-2)}{(x-2)^4} = \frac{4}{(x-2)^3}$

Step 3: Third Derivative

We now differentiate the second derivative $f''(x) = \frac{4}{(x-2)^3}$ using the chain rule again. Let $h(x) = (x-2)^3$, so the derivative of $h(x)$ is:

$h'(x) = 3(x-2)^2$

Thus, applying the chain rule:

$f'''(x) = \frac{d}{dx} \left( \frac{4}{(x-2)^3} \right) = 4 \cdot \frac{-3(x-2)^2}{(x-2)^6} = \frac{-12(x-2)^2}{(x-2)^6} = \frac{-12}{(x-2)^4}$

Final Answer:

$\frac{d^3y}{dx^3} = \frac{-12}{(x-2)^4}$

Would you like further clarification or details on any of these steps?

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Here are 5 related questions you might find helpful:

1. How do you compute higher-order derivatives for more complex functions?
2. What is the geometric interpretation of the third derivative?
3. How does the quotient rule work for rational functions in general?
4. Can you explain how the chain rule applies to implicit differentiation?
5. How can higher-order derivatives help in approximating functions?

Tip: Always simplify expressions as early as possible to avoid complex calculations later on!