Math Problem Statement

Find the derivative of the function f(x) = ³√x(x + 5).

Solution

The function given is:

f(x)=x3(x+5)f(x) = \sqrt[3]{x}(x + 5)

We are tasked with finding the derivative of this function. To do so, we will apply the product rule of differentiation and handle the cube root as a fractional exponent.

Step 1: Rewrite the function

First, express the cube root as a fractional exponent:

f(x)=x1/3(x+5)f(x) = x^{1/3}(x + 5)

Step 2: Apply the Product Rule

The product rule states:

(fg)=fg+fg(fg)' = f'g + fg'

Here, f(x)=x1/3f(x) = x^{1/3} and g(x)=x+5g(x) = x + 5.

Differentiating f(x)=x1/3f(x) = x^{1/3}:

f(x)=13x2/3f'(x) = \frac{1}{3}x^{-2/3}

Differentiating g(x)=x+5g(x) = x + 5:

g(x)=1g'(x) = 1

Step 3: Apply the Product Rule

Now substitute these into the product rule:

f(x)=(13x2/3)(x+5)+(x1/3)(1)f'(x) = \left(\frac{1}{3}x^{-2/3}\right)(x + 5) + \left(x^{1/3}\right)(1)

Step 4: Simplify the expression

f(x)=13x2/3(x+5)+x1/3f'(x) = \frac{1}{3}x^{-2/3}(x + 5) + x^{1/3}

You can leave the derivative in this form, or if needed, further simplify:

f(x)=x+53x2/3+x1/3f'(x) = \frac{x + 5}{3x^{2/3}} + x^{1/3}

This is the derivative of the function.


Do you want more details on any part of this process, or have any questions?

Here are five related questions:

  1. How does the product rule apply when both terms are functions of xx?
  2. How do you simplify expressions with fractional exponents?
  3. What is the difference between applying the product rule and the chain rule?
  4. How would the derivative change if the exponent were different?
  5. Can this derivative be applied to find the slope at a specific point?

Tip: When simplifying derivatives with fractional exponents, always rewrite them in forms that are easier to work with (e.g., as powers of xx instead of roots).

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Math Problem Analysis

Mathematical Concepts

Differentiation
Product Rule
Fractional Exponents

Formulas

Product Rule: (fg)' = f'g + fg'
Derivative of a power: d/dx[x^n] = n*x^(n-1)

Theorems

Product Rule of Differentiation

Suitable Grade Level

Grades 11-12