Understanding the Function and the Problem

The given function is \(y = \ln 2 \left(\frac{x^2−x+1}{x^2+x+1}\right)\). We are asked to find the value of the derivative, \(\frac{\text{dy}}{\text{dx}}\), at a specific point, $x = 0$. The notation \(y = \ln 2 \left(\frac{x^2−x+1}{x^2+x+1}\right)\) is most likely interpreted as the constant \(\ln 2\) multiplied by the rational function \(\frac{x^2−x+1}{x^2+x+1}\).

So, we can write the function as \(y = C \cdot f(x)\), where \(C = \ln 2\) (a constant) and \(f(x) = \frac{x^2−x+1}{x^2+x+1}\).

Applying Differentiation Rules

To find the derivative \(\frac{\text{dy}}{\text{dx}}\), we use the constant multiple rule for differentiation, which states that \(\frac{d}{dx}(C \cdot f(x)) = C \cdot \frac{d}{dx}(f(x))\).

In our case, \(\frac{\text{dy}}{\text{dx}} = \ln 2 \cdot \frac{d}{dx}\left(\frac{x^2−x+1}{x^2+x+1}\right)\).

Now, we need to find the derivative of the rational function \(f(x) = \frac{x^2−x+1}{x^2+x+1}\). We use the quotient rule, which states that if \(f(x) = \frac{u(x)}{v(x)}\), then \(f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\).

Here, let:

• \(u(x) = x^2−x+1\)
• \(v(x) = x^2+x+1\)

Find the derivatives of $u(x)$ and $v(x)$:

• \(u'(x) = \frac{d}{dx}(x^2−x+1) = 2x−1\)
• \(v'(x) = \frac{d}{dx}(x^2+x+1) = 2x+1\)

Now apply the quotient rule formula:

\(\frac{d}{dx}\left(\frac{x^2−x+1}{x^2+x+1}\right) = \frac{(2x−1)(x^2+x+1) − (x^2−x+1)(2x+1)}{(x^2+x+1)^2}\)

Let's expand the terms in the numerator:

• \((2x−1)(x^2+x+1) = 2x(x^2+x+1) − 1(x^2+x+1) = 2x^3+2x^2+2x − x^2−x−1 = 2x^3+x^2+x−1\)
• \((x^2−x+1)(2x+1) = x^2(2x+1) − x(2x+1) + 1(2x+1) = 2x^3+x^2 − 2x^2−x + 2x+1 = 2x^3−x^2+x+1\)

Subtract the second expanded term from the first to get the numerator of the derivative:

Numerator \(= (2x^3+x^2+x−1) − (2x^3−x^2+x+1)\)

Numerator \(= 2x^3+x^2+x−1 − 2x^3+x^2−x−1\)

Numerator \(= (2x^3 − 2x^3) + (x^2 + x^2) + (x − x) + (−1 − 1)\)

Numerator \(= 0 + 2x^2 + 0 − 2 = 2x^2−2\)

The derivative of the rational function is therefore:

\(f'(x) = \frac{2x^2−2}{(x^2+x+1)^2}\)

Now substitute this back into the expression for \(\frac{\text{dy}}{\text{dx}}\):

\(\frac{\text{dy}}{\text{dx}} = (\ln 2) \cdot \frac{2x^2−2}{(x^2+x+1)^2}\)

Evaluating the Derivative at x = 0

We need to find the value of \(\frac{\text{dy}}{\text{dx}}\) when $x = 0$. Substitute $x=0$ into the derivative expression:

\(\frac{\text{dy}}{\text{dx}}\Big|_{x=0} = (\ln 2) \cdot \frac{2(0)^2−2}{((0)^2+(0)+1)^2}\)

Simplify the expression:

\(\frac{\text{dy}}{\text{dx}}\Big|_{x=0} = (\ln 2) \cdot \frac{0−2}{(0+0+1)^2}\)

\(\frac{\text{dy}}{\text{dx}}\Big|_{x=0} = (\ln 2) \cdot \frac{−2}{(1)^2}\)

\(\frac{\text{dy}}{\text{dx}}\Big|_{x=0} = (\ln 2) \cdot \frac{−2}{1}\)

\(\frac{\text{dy}}{\text{dx}}\Big|_{x=0} = −2 \ln 2\)

Based on the standard rules of differentiation and interpretation of the function notation, the value of \(\frac{\text{dy}}{\text{dx}}\) at $x=0$ is −2 ln 2.

Revision Table: Key Differentiation Rules

Additional Information: Logarithms and Rational Functions

The function involves a constant factor of \(\ln 2\). The natural logarithm \(\ln x\) is the logarithm with base $e$, where \(e \approx 2.71828\). \(\ln 2\) is a specific numerical constant value, approximately $0.693$.

The core of the function is a rational function \(\frac{x^2−x+1}{x^2+x+1}\). Rational functions are ratios of polynomials. Their derivatives are typically found using the quotient rule. Evaluating a rational function at $x=0$ usually involves substituting 0 into the expression, unless the denominator becomes zero (which is not the case here, as \(0^2+0+1=1\)).

The expression \(x^2-x+1\) is always positive because its discriminant is \((-1)^2 - 4(1)(1) = 1-4 = -3 < 0\) and the leading coefficient is positive. Similarly, \(x^2+x+1\) is always positive because its discriminant is \((1)^2 - 4(1)(1) = 1-4 = -3 < 0\) and the leading coefficient is positive. Thus, the argument of any logarithm (if \(\ln 2\) were part of the argument) would be positive for all real $x$.

The derivative calculation relies on the power rule for terms like \(x^2\), $x$, and constants, combined with the quotient rule for the fraction and the constant multiple rule for \(\ln 2\).