Prove that rho = 1/(1+ phi+ ln phi) satisfies phi rho' =

Prove that $\rho = \dfrac{1}{1+\varphi+\ln\varphi}$ satisfies $\varphi\rho' = \rho\bigl(\rho\ln\varphi - 1\bigr)$.

Step 1: Differentiate $\rho$

Write $u = 1+\varphi+\ln\varphi$, so that $\rho = u^{-1}$ and

\[ u' = 1 + \frac{1}{\varphi} = \frac{\varphi+1}{\varphi}. \]

By the reciprocal rule $\bigl(u^{-1}\bigr)' = -\dfrac{u'}{u^{2}}$: $\rho' = -\frac{\varphi+1}{\varphi\,u^{2}}, \qquad u = 1+\varphi+\ln\varphi$.

Step 2: Evaluate the left-hand side

Multiplying by $\varphi$ cancels the $\varphi$ in the denominator: $\varphi\rho' = -\frac{\varphi+1}{\bigl(1+\varphi+\ln\varphi\bigr)^{2}}$.

Step 3: Evaluate the right-hand side

First simplify the bracket over the common denominator $u$: \[\rho\ln\varphi - 1 = \frac{\ln\varphi}{u} - 1 = \frac{\ln\varphi - \bigl(1+\varphi+\ln\varphi\bigr)}{u} = \frac{-(1+\varphi)}{u}.\]

The logarithms cancel exactly. Multiplying by $\rho = \dfrac1u$: \[\rho\bigl(\rho\ln\varphi-1\bigr) = \frac{1}{u}\cdot\frac{-(1+\varphi)}{u} = -\frac{1+\varphi}{u^{2}}.\]

Step 4: Compare

Both sides equal $-\dfrac{1+\varphi}{\bigl(1+\varphi+\ln\varphi\bigr)^{2}}$, so the equation holds identically.

Step 5: Verification

\[\varphi = 1:\quad \ln 1 = 0,\ u = 2 \;\Longrightarrow\; \rho = \tfrac12,\quad \rho' = -\tfrac{2}{1\cdot 4} = -\tfrac12\]

\[\text{LHS} = 1\cdot\left(-\tfrac12\right) = -\tfrac12; \qquad \text{RHS} = \tfrac12\left(\tfrac12\cdot 0 - 1\right) = -\tfrac12 \;\;\checkmark\]