Problem 10
Question
find the second derivative of the function. $$ f(x)=x \sqrt[3]{x} $$
Step-by-Step Solution
Verified Answer
The second derivative of the function \( f(x)=x \sqrt[3]{x} \) is \( f''(x) = \frac{4}{9}x^{-2/3} \).
1Step 1: Simplify the function
The function \( f(x)=x \sqrt[3]{x} \) can be simplified by writing it as \[ f(x)=x \cdot x^{1/3}=x^{1+1/3}=x^{4/3} \]
2Step 2: Find the first derivative
The rule for differentiation powers of \( x \) is \( \frac{d}{dx} x^{n} = n x^{n-1} \). Applying this rule, the first derivative of the function becomes: \[ f'(x)= \frac{4}{3}x^{4/3 - 1} = \frac{4}{3}x^{1/3} \]
3Step 3: Find the second derivative
Once again apply the power rule, to find the second derivative: \[ f''(x) = \frac{1}{3} \cdot \frac{4}{3}x^{1/3-1} = \frac{4}{9}x^{-2/3} \]
Key Concepts
Power Rule in DifferentiationUnderstanding DifferentiationSimplification of Functions
Power Rule in Differentiation
The power rule is a quick, efficient way to calculate derivatives of functions that involve powers of \( x \). It takes the form \( \frac{d}{dx} x^n = n x^{n-1} \). This means you take the exponent \( n \), bring it out front as a coefficient, and reduce the original exponent by one.
This technique is incredibly useful:
This technique is incredibly useful:
- When you have a simple power function like \( x^3 \), where applying the power rule gives \( 3x^{2} \).
- It's equally applicable to fractional powers, as seen in the exercise where \( x^{4/3} \) transitions through the power rule to \( \frac{4}{3}x^{1/3} \).
- Even negative and zero powers can be differentiated using the power rule.
Understanding Differentiation
Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. A derivative represents the rate of change of a function's output with respect to its input, essentially describing how the function's value changes as \( x \) changes.
The purpose:
With each differentiation step using rules like the power rule, you peel back a layer to explore further into the curve's dynamics. Differentiating once gives you the slope, but differentiating again gives you the concavity, indicating acceleration or deceleration of the change.
The purpose:
- To determine slopes of functions at any point.
- Find rates of change in real-world applications, such as speed in physics.
- Identify and analyze critical points where a function has maximum, minimum or saddle points.
With each differentiation step using rules like the power rule, you peel back a layer to explore further into the curve's dynamics. Differentiating once gives you the slope, but differentiating again gives you the concavity, indicating acceleration or deceleration of the change.
Simplification of Functions
Simplifying functions helps in making the differentiation process more manageable and less error-prone. By reducing complex expressions into simpler ones, not only does it make applying rules like the power rule easier, but it reduces potential calculation mistakes.
The function initially given was \( f(x) = x \sqrt[3]{x} \), where the cube root complicates straightforward differentiation.
The function initially given was \( f(x) = x \sqrt[3]{x} \), where the cube root complicates straightforward differentiation.
- Through simplification, it transforms into \( x^{4/3} \), a basic power function.
- This allows for a seamless application of differentiation rules.
- Manages fractional exponents directly without dealing with roots.
Other exercises in this chapter
Problem 10
Find \(d y / d x\) \(\frac{x y-y^{2}}{y-x}=1\)
View solution Problem 10
Find \(d y / d u, d u / d x,\) and \(d y / d x.\) $$ y=u^{3}, u=3 x^{2}-2 $$
View solution Problem 10
Find the value of the derivative of the function at the given point. State which differentiation rule you used to find the derivative. $$\begin{array}{ll}{\text
View solution Problem 11
Use a graphing utility to graph the function and find its average rate of change on the interval. Compare this rate with the instantaneous rates of change at th
View solution