Chapter 3: Problem 16
Compute \(d f\) for the given values of \(a\) and \(h\). $$ f(x)=\sqrt[3]{x} ; a=64, h=-0.1 $$
Short Answer
Expert verified
\( df = -0.0020833 \).
Step by step solution
01
Understand the problem
We are asked to compute the differential \( df \) for the function \( f(x) = \sqrt[3]{x} \) at \( x = a = 64 \) with a small change in \( x \) of \( h = -0.1 \).
02
Find the derivative of the function
The derivative \( f'(x) \) is needed to compute \( df \). For \( f(x) = \sqrt[3]{x} = x^{1/3} \), differentiate using the power rule: \[ f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}. \]
03
Evaluate the derivative at \( a = 64 \)
Substitute \( x = 64 \) into the derivative: \[ f'(64) = \frac{1}{3\sqrt[3]{64^2}}. \]Since \( \sqrt[3]{64} = 4 \), then \( 64^2 = 4096 \) and \( \sqrt[3]{4096} = 16 \), thus:\[ f'(64) = \frac{1}{3 \times 16} = \frac{1}{48}. \]
04
Compute \( df \) using the formula \( df = f'(a) \cdot h \)
Now we can use the computed derivative to find the differential:\[ df = \frac{1}{48} \times (-0.1). \]This simplifies to:\[ df = -\frac{0.1}{48} = -\frac{1}{480} = -0.0020833. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation
Differentiation is a fundamental concept in calculus that deals with finding how a function changes as its input changes. This idea of change is crucial for understanding the behavior of functions.
Differentiation helps us:
- Analyze the rate of change of quantities, such as velocity in physics or trends in economics.
- Optimize problems to find maximum or minimum values, useful in fields like engineering and economics.
Derivative of a Function
The derivative of a function represents the rate at which the function's value changes with respect to changes in its input variable. It's like asking, "How does the function respond if we make a tiny tweak to the input?" For the function \( f(x) = \sqrt[3]{x} = x^{1/3} \), its derivative \( f'(x) \) was found using rules of differentiation. In our exercise, the derivative \( f'(x) \) is calculated as:\[f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}\]The derivative tells you that for a very small change in \( x \), the change in \( f(x) \) is proportional to the derivative. At \( x = 64 \), the derivative \( f'(64) \) is computed and it informs us precisely how the function \( f(x) \) behaves around that point.
Power Rule
The Power Rule is a handy tool in calculus for differentiating expressions where the variable is raised to a power. It simplifies differentiation significantly and is expressed as:If \( f(x) = x^n \), then the derivative \( f'(x) = nx^{n-1} \).This rule was applied in our exercise for the function \( f(x) = x^{1/3} \). Using the Power Rule, the derivative of \( f(x) \) becomes:\[f'(x) = \frac{1}{3}x^{-2/3}\]This simply means that you bring down the power as a coefficient (which becomes \( \frac{1}{3} \)), and subtract one from the original power \( (1/3) - 1 = -2/3 \). This rule is particularly helpful because it saves time and avoids errors when working with polynomials or any simple power functions.