We are tasked with finding the derivative $\frac{dy}{dx}$ of the function:

$y = (x^{0.5} - 6x - 7)(x^{-1} + x^{-2})$

We will use the product rule and the power rule to differentiate this function. The product rule states that if $y = f(x)g(x)$, then:

$\frac{dy}{dx} = f'(x)g(x) + f(x)g'(x)$

Step 1: Define $f(x)$ and $g(x)$

Let:

• $f(x) = x^{0.5} - 6x - 7$
• $g(x) = x^{-1} + x^{-2}$

Step 2: Differentiate $f(x)$

The derivative of $f(x) = x^{0.5} - 6x - 7$ is:

$f'(x) = \frac{1}{2}x^{-0.5} - 6$

Step 3: Differentiate $g(x)$

The derivative of $g(x) = x^{-1} + x^{-2}$ is:

$g'(x) = -x^{-2} - 2x^{-3}$

Step 4: Apply the product rule

Now, applying the product rule:

$\frac{dy}{dx} = f'(x)g(x) + f(x)g'(x)$

Substitute $f'(x)$, $f(x)$, $g'(x)$, and $g(x)$:

$\frac{dy}{dx} = \left(\frac{1}{2}x^{-0.5} - 6\right)(x^{-1} + x^{-2}) + (x^{0.5} - 6x - 7)(-x^{-2} - 2x^{-3})$

Step 5: Simplify

Now we expand both terms:

1. For $\left(\frac{1}{2}x^{-0.5} - 6\right)(x^{-1} + x^{-2})$:

$= \frac{1}{2}x^{-0.5}x^{-1} + \frac{1}{2}x^{-0.5}x^{-2} - 6x^{-1} - 6x^{-2}$ $= \frac{1}{2}x^{-1.5} + \frac{1}{2}x^{-2.5} - 6x^{-1} - 6x^{-2}$

2. For $(x^{0.5} - 6x - 7)(-x^{-2} - 2x^{-3})$:

$= x^{0.5}(-x^{-2}) + x^{0.5}(-2x^{-3}) - 6x(-x^{-2}) - 6x(-2x^{-3}) - 7(-x^{-2}) - 7(-2x^{-3})$ $= -x^{-1.5} - 2x^{-2.5} + 6x^{-1} + 12x^{-2} + 7x^{-2} + 14x^{-3}$

Step 6: Combine all terms

Now, combine all the terms from the two expansions:

$\frac{dy}{dx} = \frac{1}{2}x^{-1.5} + \frac{1}{2}x^{-2.5} - 6x^{-1} - 6x^{-2} - x^{-1.5} - 2x^{-2.5} + 6x^{-1} + 12x^{-2} + 7x^{-2} + 14x^{-3}$

Step 7: Simplify further

Group the like terms:

$\frac{dy}{dx} = \left(\frac{1}{2}x^{-1.5} - x^{-1.5}\right) + \left(\frac{1}{2}x^{-2.5} - 2x^{-2.5}\right) + (-6x^{-1} + 6x^{-1}) + (-6x^{-2} + 12x^{-2} + 7x^{-2}) + 14x^{-3}$

$= -\frac{1}{2}x^{-1.5} - \frac{3}{2}x^{-2.5} + 13x^{-2} + 14x^{-3}$

Final Answer:

$\frac{dy}{dx} = -\frac{1}{2}x^{-1.5} - \frac{3}{2}x^{-2.5} + 13x^{-2} + 14x^{-3}$

Would you like to explore any part of this solution in more detail or have any questions?

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Here are five related questions for further exploration:

1. What is the interpretation of $\frac{dy}{dx}$ in this context?
2. How would the solution change if there were additional terms in the original expression?
3. Can you factor the result of $\frac{dy}{dx}$?
4. How would you integrate the derivative $\frac{dy}{dx}$?
5. How does the product rule apply in cases with more than two terms?

Tip: Always break down complex expressions by applying fundamental rules like the product rule and chain rule step by step.