Problem 26
Question
Find the derivative of the function \(f\) by using the rules of differentiation. \(f(x)=5 x^{4 / 3}-\frac{2}{3} x^{3 / 2}+x^{2}-3 x+1\)
Step-by-Step Solution
Verified Answer
The derivative of the function \(f(x) = 5x^{4/3} - \frac{2}{3}x^{3/2} + x^2 - 3x + 1\) is \(f'(x) = 20x^{\frac{1}{3}} - x^{\frac{1}{2}} + 2x - 3\).
1Step 1: Identify terms to differentiate individually
First, break down the function \(f(x) = 5x^{4/3} - \frac{2}{3}x^{3/2} + x^2 - 3x + 1\) into individual terms that need to be differentiated. We have:
1. \(5x^{4/3}\)
2. \(-\frac{2}{3}x^{3/2}\)
3. \(x^2\)
4. \(-3x\)
5. \(1\)
2Step 2: Apply the power rule
Apply the power rule to each term using the formula for differentiation of power functions: if \(g(x) = ax^n\), then \(g'(x) = nax^{n-1}\).
1. For \(5x^{4/3}\): \[\frac{d}{dx}\left(5x^{4/3}\right) = \frac{4}{3}(5)x^{\frac{4}{3} - 1} = 20x^{\frac{1}{3}}\]
2. For \(-\frac{2}{3}x^{3/2}\): \[\frac{d}{dx}\left(-\frac{2}{3}x^{3/2}\right) = \frac{3}{2}\left(-\frac{2}{3}\right)x^{\frac{3}{2} - 1} = -x^{\frac{1}{2}}\]
3. For \(x^2\): \[\frac{d}{dx}(x^2) = 2x^{2-1} = 2x\]
4. For \(-3x\): \[\frac{d}{dx}(-3x) = -3x^{1-1} = -3\]
5. For \(1\): \[\frac{d}{dx}(1) = 0\]
3Step 3: Combine the derivatives
Combine the derivatives of each term found in step 2 using the sum/difference rule to find the derivative of the entire function: \[f'(x) = 20x^{\frac{1}{3}} - x^{\frac{1}{2}} + 2x - 3\]
Key Concepts
FunctionsDifferentiationPower Rule
Functions
Functions are the building blocks of calculus and represent relationships between two sets of numbers or variables. A function is defined as a rule that assigns each input exactly one output. The input is often represented by the variable \(x\), and the output by \(f(x)\).
For example, in our original exercise, we have a function \(f(x)=5 x^{4 / 3}-\frac{2}{3} x^{3 / 2}+x^{2}-3 x+1\). This expression defines a rule that relates each input \(x\) to a specific output \(f(x)\), calculated by applying this formula. Functions are versatile and can model various real-world scenarios, from simple linear relationships to complex polynomial expressions.
Understanding the different types of functions and how they are constructed allows us to interpret and solve numerous problems. In calculus, functions are crucial because we often need to find rates of change, which requires calculating derivatives.
For example, in our original exercise, we have a function \(f(x)=5 x^{4 / 3}-\frac{2}{3} x^{3 / 2}+x^{2}-3 x+1\). This expression defines a rule that relates each input \(x\) to a specific output \(f(x)\), calculated by applying this formula. Functions are versatile and can model various real-world scenarios, from simple linear relationships to complex polynomial expressions.
Understanding the different types of functions and how they are constructed allows us to interpret and solve numerous problems. In calculus, functions are crucial because we often need to find rates of change, which requires calculating derivatives.
Differentiation
Differentiation is a key process in calculus, where we calculate the derivative of a function. The derivative provides us with the rate at which the function's value changes concerning its input variable, typically \(x\).
To differentiate a function, we apply rules that simplify finding the derivative, like the power rule or the product rule.
Using differentiation, we can determine several important aspects of a function:
To differentiate a function, we apply rules that simplify finding the derivative, like the power rule or the product rule.
Using differentiation, we can determine several important aspects of a function:
- The slope or gradient of a curve at any point.
- Where the function is increasing or decreasing.
- Identify maxima, minima, and points of inflection.
Power Rule
The power rule is a fundamental differentiation rule used extensively in calculus to find the derivative of functions that are polynomials or have powers. It states that, for a term of the form \(ax^n\), the derivative is given by \(nax^{n-1}\).
This is highly useful because it simplifies the differentiation process involving terms with power functions. For instance, in our given function \(f(x)\), we have several terms like \(5x^{4/3}\) and \(x^2\).
Applying the power rule allows us to derive these quickly:
This is highly useful because it simplifies the differentiation process involving terms with power functions. For instance, in our given function \(f(x)\), we have several terms like \(5x^{4/3}\) and \(x^2\).
Applying the power rule allows us to derive these quickly:
- \(\frac{d}{dx}(5x^{4/3}) = 20x^{1/3}\)
- \(\frac{d}{dx}(x^2) = 2x\)
Other exercises in this chapter
Problem 26
Find the derivative of each function. \(f(v)=\left(v^{-3}+4 v^{-2}\right)^{3}\)
View solution Problem 26
Find the derivative of each function. \(f(x)=\frac{x+1}{2 x^{2}+2 x+3}\)
View solution Problem 26
Let \(f(x)=\frac{1}{x-1}\). a. Find the derivative \(f^{\prime}\) of \(\bar{f}\) b. Find an equation of the tangent line to the curve at the point \(\left(-1,-\
View solution Problem 26
Find the indicated limit. \(\lim _{x \rightarrow-2}-3 x\)
View solution