Problem 30
Question
Differentiate each function $$ G(x)=\sqrt[3]{x^{5}+6 x} $$
Step-by-Step Solution
Verified Answer
The derivative is \( G'(x) = \frac{1}{3}(x^5 + 6x)^{-2/3} \cdot (5x^4 + 6) \).
1Step 1: Rewrite the Function
The given function is \( G(x) = \sqrt[3]{x^5 + 6x} \). To differentiate it, rewrite the function using exponent notation: \( G(x) = (x^5 + 6x)^{1/3} \). This makes it easier to apply differentiation rules.
2Step 2: Differentiate Using the Chain Rule
Apply the chain rule to differentiate the function \( G(x) = (x^5 + 6x)^{1/3} \). The chain rule states that if you have a composite function \( g(f(x)) \), then the derivative \( g'(f(x)) \) is \( g'(f(x)) \cdot f'(x) \). Here, the outer function is \( u^{1/3} \) and the inner function is \( u = x^5 + 6x \).
3Step 3: Find the Derivative of the Outer Function
Differentiate \( u^{1/3} \) with respect to \( u \). The derivative of \( u^{1/3} \) is \( \frac{1}{3}u^{-2/3} \) according to the power rule.
4Step 4: Find the Derivative of the Inner Function
Differentiate the inside function \( x^5 + 6x \) with respect to \( x \). Using the power rule, \( d/dx(x^5) = 5x^4 \), and for the linear term, \( d/dx(6x) = 6 \). So the derivative of the inner function is \( 5x^4 + 6 \).
5Step 5: Apply the Chain Rule
Combine the derivatives using the chain rule: multiply the derivative of the outer function \( \frac{1}{3}(x^5 + 6x)^{-2/3} \) by the derivative of the inner function \( 5x^4 + 6 \).
6Step 6: Simplify the Expression
The derivative \( G'(x) \) is given by: \[ G'(x) = \frac{1}{3}(x^5 + 6x)^{-2/3} \cdot (5x^4 + 6) \]. You may leave it in this form, or factor further if necessary.
Key Concepts
Chain RulePower RuleFunction Derivative
Chain Rule
The chain rule is an essential technique when dealing with composite functions. Composite functions are functions where one function is nested inside another. In this exercise, we have the function \( G(x) = (x^5 + 6x)^{1/3} \). Here, the outer function is \( u^{1/3} \), where \( u = x^5 + 6x \).
To use the chain rule effectively, follow these steps:
To use the chain rule effectively, follow these steps:
- Identify the outer and inner functions in your composite function.
- Differentiate the outer function with respect to the inner function.
- Next, differentiate the inner function with respect to \( x \).
- Finally, multiply these derivatives together to find the derivative of the composite function.
Power Rule
The power rule is fundamental in calculus, especially when differentiating polynomial functions. This rule is straightforward: if you have a function \( f(x) = x^n \), the derivative is \( f'(x) = n \cdot x^{n-1} \). This rule allows us to quickly find derivatives of powers of \( x \).
In our exercise, we apply this rule multiple times. First, to find the derivative of the inner function \( x^5 + 6x \), we apply the power rule to each term:
In our exercise, we apply this rule multiple times. First, to find the derivative of the inner function \( x^5 + 6x \), we apply the power rule to each term:
- For \( x^5 \), the derivative becomes \( 5x^4 \).
- For \( 6x \), the derivative is the constant \( 6 \).
Function Derivative
Understanding how to compute a function's derivative is crucial for analyzing how functions behave, as well as solving real-world problems involving rates of change. In this exercise, we differentiated the function \( G(x) = \sqrt[3]{x^5 + 6x} \).
Here's a simple breakdown of how we approached this:
Here's a simple breakdown of how we approached this:
- First, rewrite the function using exponent notation to facilitate differentiation. Here, that meant changing \( \sqrt[3]{x^5 + 6x} \) to \( (x^5 + 6x)^{1/3} \).
- Apply the chain rule to handle the composition of the functions, identifying and differentiating the outer and inner functions separately.
- Use the power rule to differentiate each part, making it easier to calculate the overall derivative efficiently.
- Finally, combine the derivatives to find the complete derivative, resulting in \[ G'(x) = \frac{1}{3}(x^5 + 6x)^{-2/3} \cdot (5x^4 + 6) \].
Other exercises in this chapter
Problem 30
Draw a graph that is continuous, with no corners, but not differentiable, at \(x=1\).
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The initial substitution of \(x=a\) yields the form \(0 / 0 .\) Look for ways to simplify the function algebraically, or use a table or graph to determine the l
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