Find y''_x if ln (xy) = x
Find $y''_{x}$ if $\ln\bigl(xy\bigr) = x$.
Explanation of Method
Splitting the logarithm turns the relation into two independent terms.
Step 1: Split the logarithm first
\[ \ln x+\ln y = x. \tag{1} \]
Written this way the two variables are separated, and differentiating needs no quotient rule.
Step 2: First derivative
\[ \frac1x+\frac{y'}{y} = 1 \;\Longrightarrow\; y' = y\left(1-\frac1x\right) = \frac{y\bigl(x-1\bigr)}{x}. \tag{2} \]
Step 3: Differentiate again
Product rule on (2), with $\left(\dfrac{x-1}{x}\right)' = \dfrac{x-\bigl(x-1\bigr)}{x^{2}} = \dfrac{1}{x^{2}}$: \[y'' = y'\cdot\frac{x-1}{x}+\frac{y}{x^{2}}. \tag{3}\]
Step 4: Eliminate $y'$
Substituting (2) into (3): \[y'' = \frac{y\bigl(x-1\bigr)}{x}\cdot\frac{x-1}{x}+\frac{y}{x^{2}} = \frac{y\Bigl[\bigl(x-1\bigr)^{2}+1\Bigr]}{x^{2}},\]
and $\bigl(x-1\bigr)^{2}+1 = x^{2}-2x+2$.
Step 5: Verification
\[x = 1:\ \ln y = 1 \;\Longrightarrow\; y = e \approx 2.71828\]
\[y'(1) = \frac{e\cdot0}{1} = 0; \qquad y''(1) = \frac{e\bigl(1-2+2\bigr)}{1} = e \;\;\checkmark\]
\[\text{Explicitly, (1) gives } y = \frac{e^{x}}{x}, \text{ whence } y'' = e^{x}\frac{x^{2}-2x+2}{x^{3}} = \frac{y\bigl(x^{2}-2x+2\bigr)}{x^{2}} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y'' = \dfrac{y\bigl(x^{2}-2x+2\bigr)}{x^{2}}\,}\]
Remark
Remark. The relation can be solved outright: $xy = e^{x}$, so $y = \dfrac{e^{x}}{x}$. The quadratic $x^{2}-2x+2$ has no real roots (discriminant $-4$), so $y''$ never vanishes and the curve has no inflection point — it is convex on $x>0$ and concave on $x\lt 0$, where $y\lt 0$.