Find y' for y = (1- ln x)/(1+ ln x), x > 0
Find $y'$ for $y = \dfrac{1-\ln x}{1+\ln x}$, $x \gt 0$.
Explanation of Method
A quotient in which numerator and denominator have opposite derivatives.
Step 1: Identify the parts and apply the quotient rule
For $y = \dfrac{u}{v}$ recall $y' = \dfrac{u'v - uv'}{v^{2}}$. Here
\[ u = 1-\ln x,\quad u' = -\frac{1}{x}, \qquad v = 1+\ln x,\quad v' = \frac{1}{x}. \]
Note that $u' = -v'$, which is what makes the numerator collapse.
Step 2: Substitute
\[ y' = \frac{-\dfrac{1}{x}\bigl(1+\ln x\bigr) - \bigl(1-\ln x\bigr)\cdot\dfrac{1}{x}}{\bigl(1+\ln x\bigr)^{2}}. \]
Step 3: Simplify the numerator
Factor out $-\dfrac{1}{x}$; the logarithms cancel: \[-\frac{1}{x}\Bigl[\bigl(1+\ln x\bigr) + \bigl(1-\ln x\bigr)\Bigr] = -\frac{1}{x}\cdot 2 = -\frac{2}{x}.\]
Step 4: Write the final expression
\[ y' = \frac{-\dfrac{2}{x}}{\bigl(1+\ln x\bigr)^{2}} = -\frac{2}{x\bigl(1+\ln x\bigr)^{2}}. \]
Step 5: Verification
\[x = 1:\quad \ln 1 = 0 \Rightarrow u = 1,\ u' = -1,\ v = 1,\ v' = 1 \;\Longrightarrow\; y' = \frac{(-1)(1)-(1)(1)}{1} = -2\]
\[\text{Formula:}\quad y'(1) = -\frac{2}{1\cdot 1} = -2 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = -\dfrac{2}{x\bigl(1+\ln x\bigr)^{2}}, \qquad x > 0,\ x \neq e^{-1}\,}\]
Remark
Remark. The derivative is negative on the whole domain, so the function decreases on each of the intervals $\bigl(0,e^{-1}\bigr)$ and $\bigl(e^{-1},\infty\bigr)$ separately. The excluded point $x = e^{-1}$ is where $1+\ln x = 0$ and the function has a vertical asymptote. An equivalent form is $y = \dfrac{2}{1+\ln x} - 1$.