We are tasked with finding the first and second derivatives of the function:

$y = \frac{4x^3 + 7}{x}$

Step 1: Simplify the function

First, simplify the given function by dividing each term in the numerator by $x$:

$y = \frac{4x^3}{x} + \frac{7}{x}$

This simplifies to:

$y = 4x^2 + 7x^{-1}$

Step 2: First derivative

To find the first derivative, $y'$, apply the power rule $\frac{d}{dx}[x^n] = n x^{n-1}$ to each term:

[ y' = \frac{d}{dx} [4x^2] + \frac{d}{dx} [7x^{-1}] ]

Now, differentiate each term:

$y' = 4(2x^1) + 7(-1x^{-2})$

Simplifying:

$y' = 8x - \frac{7}{x^2}$

Step 3: Second derivative

To find the second derivative, $y''$, differentiate $y' = 8x - \frac{7}{x^2}$:

[ y'' = \frac{d}{dx} [8x] - \frac{d}{dx} \left[\frac{7}{x^2}\right] ]

For $8x$, the derivative is 8.

For $-\frac{7}{x^2}$, use the power rule again:

$\frac{d}{dx} \left[-7x^{-2}\right] = -7(-2x^{-3}) = \frac{14}{x^3}$

Thus, the second derivative is:

$y'' = 8 + \frac{14}{x^3}$

Final results:

• First derivative: $y' = 8x - \frac{7}{x^2}$
• Second derivative: $y'' = 8 + \frac{14}{x^3}$

Would you like more details on any step, or have any further questions?

Here are 5 related questions to further expand:

1. How would the derivatives change if the power of $x$ in the original function were higher or lower?
2. Can you apply the product or quotient rule directly to this problem without simplifying first?
3. How does the power rule for differentiation work when applied to negative exponents?
4. What happens to the behavior of the function as $x$ approaches 0 or infinity?
5. How would you interpret the second derivative physically or geometrically?

Tip: When simplifying functions before differentiating, breaking down terms can often help make the application of basic rules easier and less prone to error.