Problem 51
Question
Differentiate with respect to the independent variable. $$ (x)=\frac{3 x^{2}-2 x+1}{2 x+1} $$
Step-by-Step Solution
Verified Answer
The derivative is \( \frac{6x^2 + 6x - 4}{(2x + 1)^2} \).
1Step 1: Identify the Function Type
The function given is a rational function, which is a division of two polynomials. The numerator is \(3x^2 - 2x + 1\) and the denominator is \(2x + 1\). The differentiation of such functions is typically done using the quotient rule.
2Step 2: Recall the Quotient Rule
The quotient rule is given by: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \). Here, \(u = 3x^2 - 2x + 1\) and \(v = 2x + 1\). We need the derivatives \(u'\) and \(v'\) to apply this rule.
3Step 3: Differentiate the Numerator
Differentiate \(u = 3x^2 - 2x + 1\) with respect to \(x\). The derivative is \(u' = 6x - 2\). This is obtained using standard derivative rules for each term separately.
4Step 4: Differentiate the Denominator
Differentiate \(v = 2x + 1\) with respect to \(x\). The derivative is \(v' = 2\). This follows directly from the linear term.
5Step 5: Apply the Quotient Rule
Substitute the derivatives \(u'\), \(v'\), \(u\), and \(v\) into the quotient rule formula: \( \left(\frac{u}{v}\right)' = \frac{(6x - 2)(2x + 1) - (3x^2 - 2x + 1)(2)}{(2x + 1)^2} \).
6Step 6: Simplify the Numerator
Expand and simplify the expression: 1. Multiply: - \((6x - 2)(2x + 1) = 12x^2 + 6x - 4x - 2 = 12x^2 + 2x - 2\) - \((3x^2 - 2x + 1)(2) = 6x^2 - 4x + 2\)2. Subtract: - \(12x^2 + 2x - 2 - (6x^2 - 4x + 2) = 6x^2 + 6x - 4\)
7Step 7: Write the Final Differentiated Expression
The derivative of the function is \( \frac{6x^2 + 6x - 4}{(2x + 1)^2} \). This comes from the simplified result of the numerator over the square of the denominator.
Key Concepts
Rational FunctionsPolynomial DifferentiationDerivative
Rational Functions
Rational functions are made up of two polynomial expressions forming a fraction. The general appearance of a rational function is: \( \frac{p(x)}{q(x)} \). Here, both \( p(x) \) and \( q(x) \) stand for polynomials which can be of any degree. The function we analyzed in the exercise is an example of a rational function, where the numerator polynomial is \( 3x^2 - 2x + 1 \) and the denominator is \( 2x + 1 \).
Working with rational functions often involves their differentiation, which can sometimes seem quite challenging due to the layered polynomial structure. Specifically, distinguishing the behaviors of the numerator and the denominator separately is key before applying any differentiation techniques, like the quotient rule. It's also crucial to ensure the denominator doesn't equal zero, as that would make the function undefined at that point.
Understanding such functions is vital as they appear frequently in calculus and are crucial in real-world applications like engineering and physics.
Working with rational functions often involves their differentiation, which can sometimes seem quite challenging due to the layered polynomial structure. Specifically, distinguishing the behaviors of the numerator and the denominator separately is key before applying any differentiation techniques, like the quotient rule. It's also crucial to ensure the denominator doesn't equal zero, as that would make the function undefined at that point.
- **Numerator** - The polynomial on top, influencing the value as it increases or decreases depending on \( x \).
- **Denominator** - The polynomial on the bottom controlling the scaling and possible restrictions to the function.
Understanding such functions is vital as they appear frequently in calculus and are crucial in real-world applications like engineering and physics.
Polynomial Differentiation
Taking derivatives of polynomials, known as polynomial differentiation, forms the backbone of calculus. It's the process of quantifying how a polynomial's value changes with a shift in the independent variable, typically \( x \). The basic rules of differentiation allow us to break down each component of the polynomial separately.
For the function's numerator \( 3x^2 - 2x + 1 \):
The denominator \( 2x + 1 \) is straightforward; the derivative of \( 2x \) is 2, and again, the constant becomes zero. Hence, the derivative of our denominator is simply 2. This step-by-step process highlights how derivative rules for basic terms like \( cx^n \) make polynomial differentiation structured and efficient.
For the function's numerator \( 3x^2 - 2x + 1 \):
- To differentiate \( 3x^2 \), bring down the power (which is 2), and multiply by the coefficient 3 to get \( 6x \).
- For \( -2x \), the power is 1, which when differentiated gives \( -2 \).
- The derivative of a constant like 1 is simply zero because constants do not change.
The denominator \( 2x + 1 \) is straightforward; the derivative of \( 2x \) is 2, and again, the constant becomes zero. Hence, the derivative of our denominator is simply 2. This step-by-step process highlights how derivative rules for basic terms like \( cx^n \) make polynomial differentiation structured and efficient.
Derivative
In calculus, a derivative represents the rate at which a function's value changes as its input changes. It is a primary tool for understanding and analyzing the behavior of functions. The concept intricately applies especially when handling rational functions, requiring accurate application of methods like the quotient rule.
The quotient rule is particularly useful in finding derivatives of rational functions. It can be expressed as: \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]. In our function, \( u \) is \( 3x^2 - 2x + 1 \) and \( v \) is \( 2x + 1 \), both are polynomials that require individual differentiation:
Substituting into the quotient rule gives: \[ \frac{(6x - 2)(2x + 1) - (3x^2 - 2x + 1)(2)}{(2x + 1)^2} \], which simplifies to the final derivative \( \frac{6x^2 + 6x - 4}{(2x + 1)^2} \).
This concrete approach allows us to keep track of both numerator and denominator changes, ensuring accuracy in evaluating function shifts as \( x \) changes. Understanding derivatives is essential to exploring more complex mathematical concepts and applications across various fields.
The quotient rule is particularly useful in finding derivatives of rational functions. It can be expressed as: \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]. In our function, \( u \) is \( 3x^2 - 2x + 1 \) and \( v \) is \( 2x + 1 \), both are polynomials that require individual differentiation:
- **\( u' \)** is found using polynomial differentiation rules, resulting in \( 6x - 2 \).
- **\( v' \)** becomes 2 – derived from the simple linear term \( 2x + 1 \).
Substituting into the quotient rule gives: \[ \frac{(6x - 2)(2x + 1) - (3x^2 - 2x + 1)(2)}{(2x + 1)^2} \], which simplifies to the final derivative \( \frac{6x^2 + 6x - 4}{(2x + 1)^2} \).
This concrete approach allows us to keep track of both numerator and denominator changes, ensuring accuracy in evaluating function shifts as \( x \) changes. Understanding derivatives is essential to exploring more complex mathematical concepts and applications across various fields.
Other exercises in this chapter
Problem 51
In Problems \(1-58\), find the derivative with respect to the independent variable. $$ h(s)=\sin ^{3} s+\cos ^{3} s $$
View solution Problem 51
Differentiate the functions in Problems 1-52 with respect to the independent variable. $$ g(r)=3^{r^{1 / 5}} $$
View solution Problem 51
Differentiate the functions with respect to the independent variable. $$ h(t)=\sin (\ln (3 t)) $$
View solution Problem 51
In Problems , find the normal line, in standard form, to \(y=\) \(f(x)\) at the indicated point. $$ y=\sqrt{3} x^{4}-2 \sqrt{3} x^{2} \text { , at } x=-\sqrt{3}
View solution