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Chapter 3: Problem 1

Find the derivative of the function. \(f(x)=x^{9 / 4}\)

Short Answer

Expert verified
The derivative is \( f'(x) = \frac{9}{4}x^{5/4} \).

Step by step solution

01

Identify the Function and Power Rule

The given function is \( f(x) = x^{9/4} \). We need to use the power rule for differentiation, which states that the derivative of \( x^n \) is \( nx^{n-1} \).
02

Apply the Power Rule

To differentiate \( f(x) = x^{9/4} \), apply the power rule: multiply the exponent by the coefficient (which is 1 here) and subtract 1 from the exponent: \( f'(x) = \frac{9}{4}x^{\frac{9}{4}-1} \).
03

Simplify the Exponent

Simplify the expression for the new exponent: \( \frac{9}{4} - 1 = \frac{9}{4} - \frac{4}{4} = \frac{5}{4} \). Substitute back into the derivative to get \( f'(x) = \frac{9}{4}x^{5/4} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule
The power rule is a fundamental rule used in calculus to find derivatives of functions of the form \(x^n\), where \(n\) is any real number. Using the power rule is straightforward and it's often one of the first rules taught when learning differentiation. The rule can be stated simply: the derivative of \(x^n\) with respect to \(x\) is \(nx^{n-1}\).
  • The exponent, \(n\), is multiplied by the coefficient of \(x\). If no coefficient is immediately visible, it is assumed to be 1.
  • Then, you subtract one from the original exponent to get the new exponent.
This makes the power rule a handy tool for easily tackling derivatives, especially as it helps to quickly handle polynomials and functions involving powers of \(x\). This is precisely what we apply when differentiating \(f(x) = x^{9/4}\).
Differentiation
Differentiation is the process of finding a derivative, which provides information about the rate at which a function is changing at any given point. In the context of our exercise, differentiating \(f(x) = x^{9/4}\) involved applying the power rule.
  • The derivative of a function is often denoted by \(f'(x)\) or \(\frac{df}{dx}\).
  • It represents the slope of the tangent line to the graph of the function at any point \(x\).
  • For functions like \(x^{9/4}\), differentiation results in another function that shows how steep the graph is at each point.
By applying the power rule, we found the derivative was \(\frac{9}{4}x^{5/4}\), indicating how \(f(x)\) changes as \(x\) changes. Differentiation helps in numerous applications, such as optimizing functions to find maxima or minima, understanding rates of change in physics, and more.
Simplifying Exponents
Simplifying exponents occurs often in calculus, especially when using the power rule for differentiation. After applying the power rule to \(x^{9/4}\), the new exponent required simplification:
  • Initially, \(9/4 - 1\) may seem complex, but rewriting 1 as \(4/4\) simplifies the subtraction.
  • Subtracting \(4/4\) from \(9/4\), the exponent becomes \(5/4\).
Simplifying exponents ensures that derivatives are concise and easy to interpret, an essential skill for solving calculus problems efficiently. The ability to manipulate exponents confidently aids in solving more complex derivatives and integrals, making it a valuable aspect of mathematical proficiency.

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Most popular questions from this chapter

Suppose \(f\) has a second derivative at \(a\), and let $$ p_{2}(x)=f(a)+f^{\prime}(a)(x-a)+\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2} $$ Prove that $$ p_{2}(a)=f(a), p_{2}^{\prime}(a)=f^{\prime}(a) \quad \text { and } p_{2}^{\prime \prime}(a)=f^{\prime \prime}(a) $$

Compute \(d f\) for the given values of \(a\) and \(h\). $$ f(x)=\sqrt{x} ; a=4, h=0.2 $$

Show that the \((n+1)\) st derivative of any polynomial of degree \(n\) is 0 . What is the \((n+2)\) nd derivative of such a polynomial?

If two bodies are a distance \(r\) apart, then the gravitational force \(F(r)\) exerted by one body on the other is given by $$ F(r)=\frac{k}{r^{2}} \quad \text { for } r>0 $$ where \(k\) is a positive constant. Suppose that as a function of time, the distance between the two bodies is given by $$ r(t)=64+48 t-16 t^{2} \quad \text { for } 0

Compute \(d f\) for the given values of \(a\) and \(h\). $$ f(x)=\sqrt[3]{x} ; a=64, h=-0.1 $$
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