1° Find (d²y)/(dx²) if b²x²+a²y² = a²b². 2° Find

1° Find $\dfrac{d^{2}y}{dx^{2}}$ if $b^{2}x^{2}+a^{2}y^{2} = a^{2}b^{2}$. 2° Find $\dfrac{d^{2}x}{dy^{2}}$ if $e^{x+y} = xy$.

Step 1: Part 1°: first derivative

\[ 2b^{2}x+2a^{2}yy' = 0 \;\Longrightarrow\; y' = -\frac{b^{2}x}{a^{2}y}. \tag{1} \]

Step 2: Part 1°: second derivative

Quotient rule on (1), then substitute (1) back in: \[y'' = -\frac{b^{2}}{a^{2}}\cdot\frac{y-xy'}{y^{2}} = -\frac{b^{2}}{a^{2}}\cdot\frac{y+\dfrac{b^{2}x^{2}}{a^{2}y}}{y^{2}} = -\frac{b^{2}}{a^{2}}\cdot\frac{a^{2}y^{2}+b^{2}x^{2}}{a^{2}y^{3}}.\]

The numerator is exactly the left side of the original equation, hence equal to $a^{2}b^{2}$: $\frac{d^{2}y}{dx^{2}} = -\dfrac{b^{4}}{a^{2}y^{3}}$.

At the top of the ellipse $(0,b)$ this gives $-\dfrac{b}{a^{2}}$, which is the curvature there since $y' = 0$.

Step 3: Part 2°: differentiate with respect to $y$

Now $x$ is the function and $y$ the variable; write $x'$ for $\dfrac{dx}{dy}$: $e^{x+y}\bigl(x'+1\bigr) = x'y+x$.

The exponential can be removed at once, because $e^{x+y} = xy$: $xy\bigl(x'+1\bigr) = x'y+x \;\Longrightarrow\; x'\bigl(xy-y\bigr) = x-xy$.

\[ x' = \frac{x\bigl(1-y\bigr)}{y\bigl(x-1\bigr)}. \tag{2} \]

Step 4: Part 2°: differentiate (2)

Quotient rule with $u = x\bigl(1-y\bigr)$ and $v = y\bigl(x-1\bigr)$, so that

\[ u' = x'\bigl(1-y\bigr)-x, \qquad v' = x'y+\bigl(x-1\bigr). \]

Grouping the $x'$ terms of $u'v-uv'$ separately from the rest: \[x'\text{-part}: \ x'\Bigl[y\bigl(x-1\bigr)\bigl(1-y\bigr)-xy\bigl(1-y\bigr)\Bigr] = x'y\bigl(1-y\bigr)\bigl[-1\bigr] = -x'y\bigl(1-y\bigr),\]

\[ \text{rest}: \ -xy\bigl(x-1\bigr)-x\bigl(1-y\bigr)\bigl(x-1\bigr) = -x\bigl(x-1\bigr)\Bigl[y+\bigl(1-y\bigr)\Bigr] = -x\bigl(x-1\bigr). \]

Step 5: Part 2°: substitute and collect

Putting (2) into $-x'y\bigl(1-y\bigr)$ gives $-\dfrac{x\bigl(1-y\bigr)^{2}}{x-1}$, so

\[ u'v-uv' = -\frac{x\bigl(1-y\bigr)^{2}}{x-1}-x\bigl(x-1\bigr) = -\frac{x\Bigl[\bigl(1-y\bigr)^{2}+\bigl(x-1\bigr)^{2}\Bigr]}{x-1}. \]

Dividing by $v^{2} = y^{2}\bigl(x-1\bigr)^{2}$: \[\frac{d^{2}x}{dy^{2}} = -\dfrac{x\Bigl[\bigl(1-y\bigr)^{2}+\bigl(x-1\bigr)^{2}\Bigr]}{y^{2}\bigl(x-1\bigr)^{3}}\]

Step 6: Verification

\[1^{\circ}:\ \text{Symbolic check with } b^{2}x^{2}+a^{2}y^{2} = a^{2}b^{2} \text{ substituted: residual } 0 \;\;\checkmark\]

\[2^{\circ}:\ \text{the symmetric point } x = y = w \text{ with } e^{w} = -w,\ w \approx -0.5671432 \text{ lies on the curve}\]

\[\text{There } x' = \frac{w\bigl(1-w\bigr)}{w\bigl(w-1\bigr)} = -1 \;\;\checkmark, \qquad x'' \approx -2.25024\]