Find y'_x for (x²)/(a²)+ (y²)/(b²)=1
Find $y'_{x}$ for $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$.
Explanation of Method
The slope of an ellipse without solving for y.
Step 1: Differentiate both sides with respect to $x$
Treat $y$ as a function of $x$; the second term needs the chain rule, contributing a factor $y'$: $\frac{2x}{a^{2}} + \frac{2y}{b^{2}}\,y' = 0$.
Step 2: Solve for $y'$
\[ \frac{2y}{b^{2}}\,y' = -\frac{2x}{a^{2}} \;\Longrightarrow\; y' = -\frac{b^{2}x}{a^{2}y}. \]
Step 3: Verification
\[a = b = 1:\quad y' = -\frac{x}{y}, \text{ the unit circle of Problem 191} \;\;\checkmark\]
\[\text{Explicitly, } y = b\sqrt{1-\frac{x^{2}}{a^{2}}} \;\Longrightarrow\; y' = \frac{-b^{2}x}{a^{2}y} \;\;\checkmark\]
\[a = 5,\ b = 3,\ (x,y) = (4, \tfrac95):\quad y' = -\frac{9\cdot 4}{25\cdot\tfrac95} = -\frac{36}{45} = -\frac45 \;\;\checkmark\]
Step 4: Where the formula fails
At the two endpoints of the major axis, $(\pm a,0)$, the denominator vanishes and the tangent is vertical — precisely where the ellipse cannot be written as a single function $y(x)$. At $(0,\pm b)$ the slope is $0$, the horizontal tangents at the ends of the minor axis.
Result
Final answer:
\[\boxed{\,y' = -\dfrac{b^{2}x}{a^{2}y}, \qquad y \neq 0\,}\]
Remark
Remark. Implicit differentiation is the right tool here because solving for $y$ forces a choice of branch (the $\pm$ of the square root), whereas the implicit formula covers both at once. Note the elegant form of the tangent line at $(x_{0},y_{0})$ that follows: $\dfrac{xx_{0}}{a^{2}}+\dfrac{yy_{0}}{b^{2}} = 1$.