Question

1. - / 5.55 points

differentiate.
$y = \frac{x + 7}{x^3 + x - 2}$
$y = \square$

Explanation:

Step1: Identify numerator and denominator

Let \( u = x + 7 \), \( v = x^3 + x - 2 \)

Step2: Compute derivatives of u and v

\( u' = 1 \), \( v' = 3x^2 + 1 \)

Step3: Apply quotient rule

\( y' = \frac{u'v - uv'}{v^2} = \frac{1 \cdot (x^3 + x - 2) - (x + 7)(3x^2 + 1)}{(x^3 + x - 2)^2} \)

Step4: Expand numerator terms

\( (x^3 + x - 2) - [3x^3 + x + 21x^2 + 7] = x^3 + x - 2 - 3x^3 - x - 21x^2 -7 \)

Step5: Simplify numerator

\( -2x^3 -21x^2 -9 \)

Answer:

\( \frac{-2x^3 - 21x^2 - 9}{(x^3 + x - 2)^2} \)