To find the derivative of the function

$f(x) = \frac{2x^3 + 3x^2 - x - 5}{\sqrt[3]{x}},$

we can use the quotient rule, which states that for functions $u(x)$ and $v(x)$:

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

In this case:

• $u(x) = 2x^3 + 3x^2 - x - 5$
• $v(x) = \sqrt[3]{x} = x^{1/3}$

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

• $u'(x) = 6x^2 + 6x - 1$ (derivative of $2x^3 + 3x^2 - x - 5$)
• $v'(x) = \frac{1}{3}x^{-2/3}$ (derivative of $x^{1/3}$)

Step 2: Apply the Quotient Rule

$f'(x) = \frac{(6x^2 + 6x - 1) \cdot x^{1/3} - (2x^3 + 3x^2 - x - 5) \cdot \frac{1}{3}x^{-2/3}}{(x^{1/3})^2}.$

Step 3: Simplify the Expression

• Simplify the numerator: $(6x^2 + 6x - 1)x^{1/3} - \frac{1}{3}(2x^3 + 3x^2 - x - 5)x^{-2/3}$
• Simplify the denominator: $(x^{1/3})^2 = x^{2/3}$

So the derivative becomes:

$f'(x) = \frac{(6x^2 + 6x - 1)x^{1/3} - \frac{1}{3}(2x^3 + 3x^2 - x - 5)x^{-2/3}}{x^{2/3}}.$

Step 4: Further Simplification

Let's combine like terms and simplify if possible:

$f'(x) = \frac{6x^{2 + 1/3} + 6x^{1 + 1/3} - x^{1/3} - \frac{1}{3}(2x^{3 - 2/3} + 3x^{2 - 2/3} - x^{1 - 2/3} - 5x^{-2/3})}{x^{2/3}}.$

This is the derivative of the given function.

Would you like a detailed step-by-step simplification for this derivative or any other questions?

Related Questions

1. What is the simplified form of the derivative $f'(x)$?
2. How do you apply the quotient rule to different types of functions?
3. What is the significance of the cube root in the denominator and how does it affect the derivative?
4. How can this derivative be used to find critical points of the function?
5. What are some common mistakes to avoid when using the quotient rule?

Tip

When dealing with derivatives involving roots or exponents, rewriting them in terms of fractional exponents can simplify the differentiation process.