$2$ . $\left.\frac{d}{d x}\left(\frac{x-\ln x}{x^{2}+1}\right)\right|_{x=1}=$

Full solution

2

\left.\frac{d}{d x}\left(\frac{x-\ln x}{x^{2}+1}\right)\right|_{x=1}=

Find Derivative: First, let's find the derivative of the function using the quotient rule, which is $(v(u') - u(v')) / v^2$ where $u = x - \ln x$ and $v = x^2 + 1$ .
Derivative of $u$ : Differentiate $u = x - \ln x$ to get $u' = 1 - (1/x)$ .
Derivative of $v$ : Differentiate $v = x^2 + 1$ to get $v' = 2x$ .
Apply Quotient Rule: Now plug $u'$ , $u$ , $v'$ , and $v$ into the quotient rule formula: $\frac{(x^2 + 1)(1 - (1/x)) - (x - \ln x)(2x)}{(x^2 + 1)^2}.$
Simplify Numerator: Simplify the numerator: $(x^2 + 1 - (\frac{x^2}{x} + \frac{1}{x})) - (2x^2 - 2x \ln x)$ .
Combine Like Terms: Combine like terms in the numerator: $x^2 + 1 - x - \frac{1}{x}$ - $2x^2 - 2x \ln x$ .
Further Simplify: Further simplify the numerator: $x^2 - x + 1 - \frac{1}{x} - 2x^2 + 2x \ln x$ .
Combine Terms: Combine the $x^2$ terms and the $x$ terms: $-x^2 - x + 1 - \frac{1}{x} + 2x \ln x$ .
Evaluate at $x=1$ : Now we have the derivative: $\frac{-x^2 - x + 1 - \frac{1}{x} + 2x \ln x}{(x^2 + 1)^2}$ .
Simplify Expression: Evaluate the derivative at $x = 1$ : $\frac{(-1^2 - 1 + 1 - \frac{1}{1} + 2 \cdot 1 \cdot \ln 1)}{(1^2 + 1)^2}$ .
Further Simplify: Simplify the expression: $(-1 - 1 + 1 - 1 + 0) / (1 + 1)^2$ .
Final Simplification: Further simplify: $(-2 + 1 - 1) / 4$ .
Final Answer: Final simplification: $-2 / 4$ .
Final Answer: Final simplification: $-2 / 4$ .The final answer is $-1/2$ .