Problem 24 - Brief Applied Calculus

Question

Find the derivative of each function by using the Product Rule. Simplify your answers. $$ f(z)=(\sqrt[4]{z}+\sqrt{z})(\sqrt[4]{z}-\sqrt{z}) $$

Step-by-Step Solution

Answer

The derivative of the function is 0.

Step 1: Identify the functions

The given function is \( f(z) = (\sqrt[4]{z} + \sqrt{z})(\sqrt[4]{z} - \sqrt{z}) \). We will label \( u(z) = \sqrt[4]{z} + \sqrt{z} \) and \( v(z) = \sqrt[4]{z} - \sqrt{z} \) for use in the product rule.

Step 2: Write the Product Rule

The product rule for derivatives states that if you have a product of two functions \( u(z) \) and \( v(z) \), the derivative \( f'(z) \) is given by \( f'(z) = u'(z)v(z) + u(z)v'(z) \).

Step 3: Calculate Derivatives of u(z) and v(z)

Find \( u'(z) \) and \( v'(z) \). For \( u(z) = \sqrt[4]{z} + \sqrt{z} \): - \( \sqrt[4]{z} = z^{1/4} \) and its derivative is \( \frac{1}{4}z^{-3/4} \). - \( \sqrt{z} = z^{1/2} \) and its derivative is \( \frac{1}{2}z^{-1/2} \). Thus, \( u'(z) = \frac{1}{4}z^{-3/4} + \frac{1}{2}z^{-1/2} \).For \( v(z) = \sqrt[4]{z} - \sqrt{z} \):- The derivatives are the same in magnitude as for \( u(z) \), but with a negative sign for the second term.Thus, \( v'(z) = \frac{1}{4}z^{-3/4} - \frac{1}{2}z^{-1/2} \).

Step 4: Apply the Product Rule

Substitute \( u(z) \), \( v(z) \), \( u'(z) \), and \( v'(z) \) into the product rule: \[ f'(z) = \left( \frac{1}{4}z^{-3/4} + \frac{1}{2}z^{-1/2} \right)(\sqrt[4]{z} - \sqrt{z}) + (\sqrt[4]{z} + \sqrt{z})\left( \frac{1}{4}z^{-3/4} - \frac{1}{2}z^{-1/2} \right) \].

Step 5: Simplify the Expression

Simplify the expression from Step 4. First, distribute each derivative term:- \( \frac{1}{4}z^{-3/4}(\sqrt[4]{z} - \sqrt{z}) = \frac{1}{4}z^{-3/4}z^{1/4} - \frac{1}{4}z^{-3/4}z^{1/2} \). Simplifying gives \( \frac{1}{4}z^{-1/2} - \frac{1}{4}z^{-1/4} \).- \( \frac{1}{2}z^{-1/2}(\sqrt[4]{z} - \sqrt{z}) = \frac{1}{2}z^{-1/2}z^{1/4} - \frac{1}{2}z^{-1/2}z^{1/2} \). Simplifying gives \( \frac{1}{2}z^{-1/4} - \frac{1}{2} \).- \( \frac{1}{4}z^{-3/4}(\sqrt[4]{z} + \sqrt{z}) = \frac{1}{4}z^{-1/2} + \frac{1}{4}z^{-1/4} \).- \( -\frac{1}{2}z^{-1/2}(\sqrt[4]{z} + \sqrt{z}) = -\frac{1}{2}z^{-1/4} - \frac{1}{2} \).Adding all these results, notice that most terms cancel out:\[ \frac{1}{4}z^{-1/2} - \frac{1}{4}z^{-1/4} + \frac{1}{2}z^{-1/4} - \frac{1}{2} + \frac{1}{4}z^{-1/2} + \frac{1}{4}z^{-1/4} - \frac{1}{2}z^{-1/4} - \frac{1}{2} \] = \(-2\cdot\frac{1}{2}\) which simplifies to 0.Thus, \( f'(z) = 0 \).

Step 6: Conclusion

The derivative of the function \( f(z) = (\sqrt[4]{z} + \sqrt{z})(\sqrt[4]{z} - \sqrt{z}) \) using the product rule is \( f'(z) = 0 \). This verifies that the original function is a constant since its derivative equals zero.

Key Concepts

Derivative

In calculus, understanding how to find the derivative of a function is crucial. A derivative represents the rate of change or slope of a function at any given point. It helps us understand how a function behaves and how its output changes as its input varies.
To find the derivative of complex functions involving multiplication, we often use the product rule. The product rule is a formula that allows us to differentiate functions that are products of two simpler functions. If you have a function that is the product of two other functions, say, \( u(z) \) and \( v(z) \), then the derivative \( f'(z) \) is calculated as follows:

• Calculate the derivative of the first function \( u'(z) \)
• Multiply \( u'(z) \) by the second function \( v(z) \)
• Then, calculate the derivative of the second function \( v'(z) \)
• Multiply \( v'(z) \) by the first function \( u(z) \)
• Add these two results together to get \( f'(z) = u'(z)v(z) + u(z)v'(z) \)

This method ensures that even if functions seem complicated, we can handle their derivatives efficiently.

Simplification of Expressions

Simplification plays a critical role in the process of solving calculus problems, making it easier to interpret and verify solutions. After applying the product rule and finding an expression for a derivative, it's often useful to simplify the result as much as possible.
Simplifying an expression involves:

• Combining like terms
• Factoring expressions when possible
• Canceling terms that appear on both sides of an equation
• Using algebraic identities to make expression forms more manageable

In our problem, simplifying the expression after using the product rule led to the realization that many terms canceled out. This indicated that the result was a constant, showing significant insight into the behavior of the function itself.

Constant Function

A constant function is a function whose output value is the same regardless of the input value. Mathematically, its derivative is always zero because the rate of change of a constant value is zero. This concept is a fundamental aspect in calculus, providing clues about a function's behavior.
In this exercise, we saw that the derivative \( f'(z) = 0 \), indicating that the function \( f(z) = (\sqrt[4]{z} + \sqrt{z})(\sqrt[4]{z} - \sqrt{z}) \) is constant. Whenever you derive an expression and get zero, you can conclude that the original function is constant over its domain.
Recognizing a constant function can be powerful because it tells us the function does not change over its entire range, simplifying analysis and helping predict outcomes in various applications of calculus.