Problem 68

Question

Use the Chain Rule to differentiate each function. You may need to apply the rule more than once. $$ f(x)=\left(-x^{5}+4 x+\sqrt{2 x+1}\right)^{3} $$

Step-by-Step Solution

Verified
Answer
The derivative is \( \frac{df}{dx} = 3(-x^5 + 4x + \sqrt{2x+1})^2 \cdot (-5x^4 + 4 + (2x+1)^{-1/2}) \).
1Step 1: Identify the Functions
In the function \( f(x) = \left( -x^5 + 4x + \sqrt{2x+1} \right)^3 \), identify the outer function, which is the power function \( u^3 \), and the inner function \( u = -x^5 + 4x + \sqrt{2x+1} \). This will guide the application of the chain rule.
2Step 2: Differentiate the Outer Function
Differentiate the outer function \( u^3 \) with respect to \( u \). Using the power rule, we find \( \frac{d}{du} (u^3) = 3u^2 \).
3Step 3: Differentiate the Inner Function
To differentiate the inner function \( u = -x^5 + 4x + \sqrt{2x+1} \), use the sum rule: \( \frac{d}{dx}(-x^5) = -5x^4 \), \( \frac{d}{dx}(4x) = 4 \), and apply the chain rule to \( \sqrt{2x+1} = (2x+1)^{1/2} \), giving \( \frac{1}{2}(2x+1)^{-1/2} \cdot 2 = (2x+1)^{-1/2} \).
4Step 4: Combine Derivatives Using Chain Rule
Apply the chain rule: \( \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \). Substitute \( \frac{df}{du} = 3u^2 \) and calculate \( \frac{du}{dx} = -5x^4 + 4 + (2x+1)^{-1/2} \).
5Step 5: Substitute and Simplify
Substitute \( u = -x^5 + 4x + \sqrt{2x+1} \) back into the equation: \( \frac{df}{dx} = 3(-x^5 + 4x + \sqrt{2x+1})^2 \cdot (-5x^4 + 4 + (2x+1)^{-1/2}) \).

Key Concepts

DifferentiationPower RuleInternal and External Functions
Differentiation
Differentiation is a fundamental concept in calculus that focuses on finding how a function changes as its input changes.
It involves finding the derivative of a function, which is a new function that gives the rate of change of the original function.

For instance, if you consider a function that describes how distance changes over time, its derivative would tell you the speed at which the distance is changing.
  • The derivative is represented by \( \frac{d}{dx} \), where \( x \) is the variable.
  • Differentiation allows us to determine slopes of curves, optimize processes, and solve various real-life problems.
In our example, we are using the chain rule for differentiation.
This allows us to differentiate composite functions, or functions within other functions. By doing this, we can tackle more complex equations and extract meaningful information about their behavior.
Power Rule
The power rule is one of the simplest rules in differentiation, making it very popular in calculating derivatives.
It states that if you have a function in the form of \( x^n \), where \( n \) is a constant, its derivative is \( nx^{n-1} \).

For example, differentiating \( x^3 \) would give you \( 3x^2 \).
  • The power rule efficiently handles simple polynomial expressions.
  • Remember: the exponent decreases by one, and you multiply by the original exponent.
In the provided exercise, when we differentiated the outer function \( u^3 \), we applied the power rule, resulting in \( 3u^2 \).
By mastering the power rule, students can quickly find derivatives of many functions, which is essential for more advanced calculus operations.
Internal and External Functions
In a composite function, understanding internal and external functions is crucial for applying differentiation rules like the chain rule.
The internal function is the one nested within another function, while the external function operates on the entire composition.

For the function \( f(x) = (-x^5 + 4x + \sqrt{2x+1})^3 \), the internal function is \( u = -x^5 + 4x + \sqrt{2x+1} \) and the external function is \( u^3 \).
  • Distinguishing these helps in correctly applying the chain rule.
  • Start by differentiating the external function, and follow by differentiating the internal function.
The internal function may itself require additional differentiation rules or nested chain rule application.
In our example, differentiating \( \sqrt{2x+1} \) used the chain rule again.
This method is powerful for untangling layers of functions to find their derivatives accurately.