Q.If $x^2 + y^2 = 25$, find $\dfrac{dy}{dx}$ by implicit differentiation, and its value at the point $(3, 4)$.

The equation is not solved for $y$, so differentiate both sides w.r.t. $x$ treating $y$ as a function of $x$.

The relation $x^2 + y^2 = 25$ (a circle) mixes $x$ and $y$, so use implicit differentiation (§6): differentiate every term with respect to $x$, treating $y$ as a function of $x$.

Differentiate term by term. $\dfrac{d}{dx}(x^2) = 2x$; $\dfrac{d}{dx}(y^2) = 2y\dfrac{dy}{dx}$ (chain rule — $y$ depends on $x$); $\dfrac{d}{dx}(25) = 0$. So

Solve for $\dfrac{dy}{dx}$. $2y\dfrac{dy}{dx} = -2x$, hence $\dfrac{dy}{dx} = -\dfrac{x}{y}$.

Evaluate at $(3, 4)$. $\dfrac{dy}{dx} = -\dfrac{3}{4}$. …

The defining error is differentiating $y^2$ as $2y$ instead of $2y\tfrac{dy}{dx}$ — forgetting that $y$ is a function of $x$. Remember every $y$-term gains a $\tfrac{dy}{dx}$ factor. It is als …