Given that: \(\frac{d}{{dx}}\left( {\frac{{1 + {x^2} + {x^4}}}{{1 + x + {x^2}}}} \right) = Ax + B.\)
What is the value of B?
-1
Understanding the Derivative Problem
The question asks us to find the value of the constant \(B\) given the equation involving a derivative: \(\frac{d}{{dx}}\left( {\frac{{1 + {x^2} + {x^4}}}{{1 + x + {x^2}}}} \right) = Ax + B\). To find \(B\), we first need to calculate the derivative on the left side of the equation. This involves simplifying the expression inside the derivative before performing the differentiation.
Simplifying the Polynomial Expression
We are given the expression \(\frac{{1 + {x^2} + {x^4}}}{{1 + x + {x^2}}}\). We need to simplify this fraction. Let's look at the numerator, \(1 + x^2 + x^4\). This is a standard form that can be factored. We can rewrite it as \((1 + x^2)^2 - x^2\). This is a difference of squares, which factors as \((a-b)(a+b)\) where \(a = 1 + x^2\) and \(b = x\).
So, \(1 + x^2 + x^4 = (1 + x^2 - x)(1 + x^2 + x)\).
Now, substitute this back into the fraction:
\(\frac{{1 + {x^2} + {x^4}}}{{1 + x + {x^2}}} = \frac{{(1 - x + {x^2})(1 + x + {x^2})}}{{1 + x + {x^2}}}\)
Assuming \(1 + x + x^2 \neq 0\), we can cancel out the common term \((1 + x + x^2)\) from the numerator and the denominator.
The simplified expression is \(1 - x + x^2\).
Calculating the Derivative
Now that we have simplified the expression, we need to find its derivative with respect to \(x\):
\(\frac{d}{{dx}}\left( {1 - x + {x^2}} \right)\)
Using the rules of differentiation (derivative of a constant is 0, derivative of \(x\) is 1, and derivative of \(x^n\) is \(nx^{n-1}\)):
- \(\frac{d}{{dx}}(1) = 0\)
- \(\frac{d}{{dx}}(-x) = -1\)
- \(\frac{d}{{dx}}(x^2) = 2x^{2-1} = 2x\)
So, the derivative is:
\(\frac{d}{{dx}}(1 - x + x^2) = 0 - 1 + 2x = 2x - 1\)
Finding the Value of B
We are given that the derivative is equal to \(Ax + B\).
We found the derivative to be \(2x - 1\).
So, we have the equation:
\(2x - 1 = Ax + B\)
To find the values of \(A\) and \(B\), we can compare the coefficients of the corresponding terms on both sides of the equation.
- Comparing the coefficients of \(x\): On the left side, the coefficient of \(x\) is 2. On the right side, the coefficient of \(x\) is \(A\). Therefore, \(A = 2\).
- Comparing the constant terms: On the left side, the constant term is -1. On the right side, the constant term is \(B\). Therefore, \(B = -1\).
The value of \(B\) is -1.
Summary of Steps
- Simplify the given rational expression \(\frac{{1 + {x^2} + {x^4}}}{{1 + x + {x^2}}}\) by factoring the numerator and cancelling the common term.
- Differentiate the simplified expression with respect to \(x\).
- Equate the resulting derivative polynomial to \(Ax + B\).
- Compare the constant terms on both sides of the equation to find the value of \(B\).
| Step | Calculation | Result |
|---|---|---|
| Simplify Expression | \(\frac{{1 + {x^2} + {x^4}}}{{1 + x + {x^2}}} = \frac{{(1 - x + {x^2})(1 + x + {x^2})}}{{1 + x + {x^2}}}\) | \(1 - x + x^2\) |
| Calculate Derivative | \(\frac{d}{{dx}}(1 - x + x^2)\) | \(2x - 1\) |
| Compare with \(Ax+B\) | \(2x - 1 = Ax + B\) | \(A=2, B=-1\) |
Thus, the value of \(B\) is -1.
Revision Table: Derivative and Polynomial Concepts
| Concept | Description |
|---|---|
| Derivative | Measures the instantaneous rate of change of a function. Notation: \(\frac{dy}{dx}\) or \(f'(x)\). |
| Polynomial | An expression consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. Example: \(ax^2 + bx + c\). |
| Polynomial Factorization | Breaking down a polynomial into a product of simpler polynomials. Example: \(x^2 - y^2 = (x-y)(x+y)\). |
| Power Rule of Differentiation | If \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\). |
| Sum/Difference Rule of Differentiation | \(\frac{d}{dx}[f(x) \pm g(x)] = \frac{d}{dx}[f(x)] \pm \frac{d}{dx}[g(x)]\). |
Additional Information: Related Calculus and Algebra
Understanding polynomial factorization is key to simplifying expressions in calculus problems. The identity \(1 + x^2 + x^4 = (1 - x + x^2)(1 + x + x^2)\) is particularly useful and derived from the difference of squares formula applied to \((1 + x^2)^2 - x^2\). This technique of simplifying rational functions before differentiating is often much easier than using the quotient rule directly on the original expression.
Comparing coefficients is a standard algebraic technique used when two polynomial expressions are equal. If \(P(x) = Q(x)\) for all \(x\), where \(P(x)\) and \(Q(x)\) are polynomials, then the coefficients of corresponding powers of \(x\) in \(P(x)\) and \(Q(x)\) must be equal. In our problem, \(2x - 1 = Ax + B\) is an equality of two linear polynomials, allowing us to equate the coefficients of \(x\) and the constant terms.
Important Questions from Evaluation of derivatives
The derivative of In(x + sin x) with respect to (x + cos x) is
If x ay b= (x - y) a+b , then the value of \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} - \frac{{\rm{y}}}{{\rm{x}}}\) is equal to
Let f(x + y) = f(x) f(y) for all x and y. Then what is f’(5) equal to [where f’(x) is the derivative of f(x)]?
What f’(x) equal to when 0 < x < 1?
Which of the following equations is/are correct?
1. f(-2) = f(5)
2. f”(-2) + f”(0.5) + f”(3) = 4
Select the correct answer using the code given below:
