To differentiate the function $y = \frac{x + 5}{x^3 + x - 1}$, we'll use the quotient rule.

Quotient Rule:

If you have a function of the form $y = \frac{u(x)}{v(x)}$, the derivative is:

$y' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

Where $u(x) = x + 5$ and $v(x) = x^3 + x - 1$.

Step 1: Differentiate $u(x)$ and $v(x)$

• $u(x) = x + 5$, so $u'(x) = 1$.
• $v(x) = x^3 + x - 1$, so $v'(x) = 3x^2 + 1$.

Step 2: Apply the quotient rule

Now substitute $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$ into the quotient rule formula:

$y' = \frac{(1)(x^3 + x - 1) - (x + 5)(3x^2 + 1)}{(x^3 + x - 1)^2}$

Step 3: Simplify the numerator

First, expand the terms in the numerator:

$1 \cdot (x^3 + x - 1) = x^3 + x - 1$ $(x + 5) \cdot (3x^2 + 1) = (x \cdot 3x^2 + x \cdot 1) + (5 \cdot 3x^2 + 5 \cdot 1) = 3x^3 + x + 15x^2 + 5$

Now substitute back into the numerator:

$y' = \frac{x^3 + x - 1 - (3x^3 + x + 15x^2 + 5)}{(x^3 + x - 1)^2}$

Simplify the terms:

$y' = \frac{x^3 + x - 1 - 3x^3 - x - 15x^2 - 5}{(x^3 + x - 1)^2}$

$y' = \frac{-2x^3 - 15x^2 - 6}{(x^3 + x - 1)^2}$

So, the derivative of $y = \frac{x + 5}{x^3 + x - 1}$ is:

$y' = \frac{-2x^3 - 15x^2 - 6}{(x^3 + x - 1)^2}$

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Would you like more details on this process, or do you have another question?

Here are some related questions:

1. How do you apply the product rule in differentiation?
2. What happens when you differentiate a function that has a power of a quotient?
3. How do you handle higher-order derivatives of rational functions?
4. What is the chain rule, and when do you use it?
5. How do you find critical points and use them to determine concavity?

Tip: When differentiating complex rational functions, always double-check your algebra when simplifying the numerator and denominator!