Differentiating implicitly AB & BC

On the circle $x^2 + y^2 = 25$, the value of $y$ depends on where you are in $x$. So $y$ is a function of $x$, even though no formula for it has been written down. That single fact is what makes $y^2$ a composition: it is the squaring machine applied to the function $y(x)$.

The chain rule then applies exactly as it always does. Differentiating $y^2$ with respect to $x$ gives the outer derivative $2y$ times the derivative of the inside, which is $\frac{dy}{dx}$:

$$\frac{d}{dx}\left[y^2\right] = 2y\frac{dy}{dx}.$$

Writing $2y$ alone is the defining error of this topic. It treats $y$ as if it were the variable of differentiation, or as if it were a constant, and both readings lose the same factor. The same applies to every function of $y$: $\frac{d}{dx}\sin y = \cos y\frac{dy}{dx}$, and $\frac{d}{dx}e^y = e^y\frac{dy}{dx}$.

Terms in $x$ alone behave normally, because $\frac{dx}{dx} = 1$. That asymmetry is the point: $x$-terms are plain, $y$-terms carry a factor.

A term such as $xy$ contains both variables multiplied together, so before the chain rule can be applied to the $y$ the product rule has to be applied to the term. Both rules, in that order:

$$\frac{d}{dx}\left[xy\right] = 1\cdot y + x\cdot\frac{dy}{dx} = y + x\frac{dy}{dx}.$$

Skipping the product rule and writing $\frac{dy}{dx}$, or writing $x\frac{dy}{dx}$ alone, drops the $y$ that comes from differentiating the $x$ factor. A heavier example makes the layering explicit:

$$\frac{d}{dx}\left[x^2y^3\right] = 2xy^3 + x^2\cdot 3y^2\frac{dy}{dx}.$$

The first piece differentiates $x^2$ and leaves $y^3$ alone. The second leaves $x^2$ alone and differentiates $y^3$, which is itself a composition and so contributes $3y^2\frac{dy}{dx}$.

After differentiating, $\frac{dy}{dx}$ appears in several places and the rest is algebra:

1. Differentiate both sides with respect to $x$, tagging every $y$-term with $\frac{dy}{dx}$.
2. Collect every term containing $\frac{dy}{dx}$ on one side, everything else on the other.
3. Factor $\frac{dy}{dx}$ out of that side.
4. Divide.

For $x^2 + y^2 = 25$: differentiating gives $2x + 2y\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{x}{y}$. The answer depends on both coordinates, which is normal for an implicit derivative and is the reason a point must be given in full. At $(3, 4)$ the slope is $-\frac{3}{4}$; at $(3, -4)$ the same $x$ gives $+\frac{3}{4}$.

Two ordering errors are worth naming. Dividing before collecting strands a $\frac{dy}{dx}$ on the wrong side. And substituting the point into the original equation before differentiating turns the equation into a numerical statement whose derivative is $0 = 0$.

The formula $\frac{dy}{dx} = -\frac{x}{y}$ describes the circle's tangent everywhere at once, and it stays honest at the awkward places. Where $y = 0$, at $(\pm 5, 0)$, the expression is undefined, which is exactly right: the tangent there is vertical.

Implicit differentiation also handles curves that fail the vertical line test, which is why it exists. A curve like $x^3 + y^3 = 6xy$ has no clean solved form at all, yet its slope at any point on it is one differentiation and a line of algebra away. Solving for $y$ first is not merely harder here; it is not available.