To solve the given differentiation problems in Examples 1-5, we will use the quotient rule for derivatives.

The quotient rule states:

$\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2},$ where $u(x)$ is the numerator and $v(x)$ is the denominator.

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Solution for Example 1

Differentiate:

$\frac{d}{dx} \left( \frac{x^2 + 3x}{x^3 - 2x} \right).$

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

Using the quotient rule:

$\frac{d}{dx} \left( \frac{x^2 + 3x}{x^3 - 2x} \right) = \frac{(2x + 3)(x^3 - 2x) - (x^2 + 3x)(3x^2 - 2)}{(x^3 - 2x)^2}.$

Simplify the numerator:

1. Expand $(2x + 3)(x^3 - 2x)$: $(2x + 3)(x^3 - 2x) = 2x^4 - 4x^2 + 3x^3 - 6x.$

2. Expand $(x^2 + 3x)(3x^2 - 2)$: $(x^2 + 3x)(3x^2 - 2) = 3x^4 - 2x^2 + 9x^3 - 6x.$

Now subtract the two results:

$\text{Numerator} = \left( 2x^4 - 4x^2 + 3x^3 - 6x \right) - \left( 3x^4 - 2x^2 + 9x^3 - 6x \right).$

Combine like terms:

[ \text{Numerator} = 2x^4 - 3x^4 - 4x^2 + 2x^2 + 3x^3 - 9x^3 - 6x + 6x