Problem 29
Question
Differentiate $$ f(x)=a(x+1)(2 x-1) $$ with respect to \(x\). Assume that \(a\) is a positive constant.
Step-by-Step Solution
Verified Answer
The derivative of the function is \(f'(x) = a(4x + 1)\).
1Step 1: Identify the Product Rule
To differentiate the function \(f(x) = a(x+1)(2x-1)\), we need to apply the product rule. Recall that the product rule states: if \(u(x)\) and \(v(x)\) are functions of \(x\), then the derivative \((uv)' = u'v + uv'\). Here, let \(u(x) = x+1\) and \(v(x) = 2x - 1\).
2Step 2: Differentiate Each Function
Differentiate \(u(x)\) and \(v(x)\):- \(u(x) = x+1\) so \(u'(x) = 1\) since the derivative of \(x\) is \(1\) and the derivative of a constant is \(0\).- \(v(x) = 2x-1\) so \(v'(x) = 2\) since the derivative of \(2x\) is \(2\) and the derivative of a constant is \(0\).
3Step 3: Apply the Product Rule
Substitute \(u(x)\), \(v(x)\), \(u'(x)\), and \(v'(x)\) into the product rule formula: \((x+1)'(2x-1) + (x+1)(2x-1)' = 1(2x-1) + (x+1)(2)\).
4Step 4: Simplify the Expression
Simplify the expression: \(1(2x-1) + 2(x+1) = 2x - 1 + 2x + 2\).Combine like terms: \(2x + 2x - 1 + 2 = 4x + 1\).
5Step 5: Account for the Constant 'a'
Since the function \(f(x)\) is multiplied by the constant \(a\), the derivative is the constant times the derivative of the function: \(f'(x) = a(4x+1)\).
6Step 6: Final Step: Conclusion
Thus, the derivative of \(f(x) = a(x+1)(2x-1)\) with respect to \(x\) is:\(f'(x) = a(4x + 1)\).
Key Concepts
Product RulePolynomial FunctionsDerivative of a Constant
Product Rule
When we have a function that is a product of two smaller functions, the product rule is our go-to method for differentiation. This rule helps us find the derivative of the product quickly and accurately. The general formula for the product rule is \((uv)' = u'v + uv'\).
This means if you have functions \(u(x)\) and \(v(x)\), you'll differentiate each separately to find \(u'(x)\) and \(v'(x)\). Once you have these, the next step is simple. Just plug them into the product rule formula.
This means if you have functions \(u(x)\) and \(v(x)\), you'll differentiate each separately to find \(u'(x)\) and \(v'(x)\). Once you have these, the next step is simple. Just plug them into the product rule formula.
- Identify the two functions: Break down your main function into two parts that are multiplied.
- Differentiate: Find the derivatives of each of those parts.
- Apply the formula: Use \(u'v + uv'\) to assemble your answer.
Polynomial Functions
Polynomial functions are expressions that consist of variables and coefficients, usually combined using addition, subtraction, and multiplication. These functions are easy to work with due to their straightforward structure. An important fact about polynomials is that they have different terms with varying powers of \(x\).
Standard polynomial functions look like \(a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0\). Here, each \(a_i\) represents a constant, and each \(x^i\) raises \(x\) to some power.
Standard polynomial functions look like \(a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0\). Here, each \(a_i\) represents a constant, and each \(x^i\) raises \(x\) to some power.
- Degree: The highest power of \(x\) is called the degree of the polynomial.
- Coefficients: These are the numerical factors multiplying the powers of \(x\).
- Easy differentiation: Just multiply each term by its power and decrease the power by 1.
Derivative of a Constant
Differentiating constants is one of the simplest tasks in calculus. Constants are terms with no \(x\) variable attached, such as just a numerical value. The rule here is straightforward: the derivative of a constant is always zero.
Think of this as a flat line on a graph, where things don't change. Since there's no slope or move in any direction (up or down), the change rate is zero.
In mathematical terms, if \(c\) is a constant, then \(\frac{d}{dx}(c) = 0\).
Think of this as a flat line on a graph, where things don't change. Since there's no slope or move in any direction (up or down), the change rate is zero.
In mathematical terms, if \(c\) is a constant, then \(\frac{d}{dx}(c) = 0\).
- Flat line: Constant results don't rise or fall – they remain consistent.
- No change: Since there's no variable interaction, there's nothing to differentiate.
- Simple application: Just remember – see a constant, derivative is zero.
Other exercises in this chapter
Problem 29
Find the derivative with respect to the independent variable. $$ f(x)=\sqrt{\sin \left(2 x^{2}-1\right)} $$
View solution Problem 29
Differentiate the functions with respect to the independent variable. (Note that log denotes the logarithm to base 10.) $$ f(x)=\ln \left(2 x^{3}-x\right) $$
View solution Problem 29
Differentiate $$ f(x)=r s^{2} x^{3}-r x+s $$ with respect to \(x\). Assume that \(r\) and \(s\) are constants.
View solution Problem 29
In Problems , graph each function and, on the basis of the graph, guess where the function is not differentiable. (Assume the largest possible domain.)$$ y=2-|x
View solution