Implicit differentiation

$x^2 + y^2 = 25$ is a circle, and no function describes it — a circle fails the vertical line test. Yet it plainly has a tangent line at every point, so a derivative exists locally even though $y$ is not a function of $x$ globally.

Implicit differentiation gets that derivative without ever solving for $y$.

The one rule

Differentiate both sides with respect to $x$, treating $y$ as a function of $x$. Every $y$ you differentiate produces a factor of $\tfrac{dy}{dx}$, by the chain rule.

For $x^2 + y^2 = 25$:

$$2x + 2y\frac{dy}{dx} = 0 \qquad\Longrightarrow\qquad \frac{dy}{dx} = -\frac{x}{y}$$

The second line is where the $\tfrac{dy}{dx}$ comes from: differentiating $y^2$ with respect to $y$ gives $2y$, and the chain rule multiplies by $\tfrac{dy}{dx}$ to convert it to a derivative with respect to $x$.

Checking against the explicit route

The circle is one of the rare cases where you can solve for $y$, which makes it a good check. The upper half is $y = \sqrt{25 - x^2}$:

That is $-\tfrac{x}{y}$, since $y = \sqrt{25-x^2}$ on the upper half. The two routes agree, and the implicit one gave the answer in one line without a square root and without splitting the circle into halves.

That economy is the point. For $x^3 + y^3 = 6xy$ — the folium of Descartes — solving for $y$ means a cubic formula. Differentiating implicitly takes a line:

$$3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$$

Collect the $\tfrac{dy}{dx}$ terms on one side and factor:

$$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}$$

The answer depends on both $x$ and $y$, which is normal for an implicit derivative and is exactly right — a curve that doubles back has different slopes at two points with the same $x$.

Products need the product rule

A term with both variables needs the product rule and the chain rule:

$$\frac{d}{dx}(xy) = 1\cdot y + x\frac{dy}{dx}$$

That verified line is the first half — the derivative of $xy$ treating $y$ as constant is $y$ — and the second half, $x\tfrac{dy}{dx}$, is what implicit differentiation adds. Missing it is the same omission as before, wearing a different hat.

Tangent lines

The usual exam question: find the tangent to $x^2 + y^2 = 25$ at $(3, 4)$.

$\tfrac{dy}{dx} = -\tfrac{x}{y} = -\tfrac{3}{4}$, so the tangent is

$$y - 4 = -\tfrac34(x - 3)$$

Sanity check: the radius to $(3,4)$ has slope $\tfrac43$, and $-\tfrac34$ is its negative reciprocal — the tangent is perpendicular to the radius, as a circle’s tangent must be. Geometry confirming calculus is a free check, and worth taking.

Inverse functions

Implicit differentiation gives the inverse-function rule immediately. If $y = f^{-1}(x)$ then $f(y) = x$, and differentiating both sides:

$$f'(y)\frac{dy}{dx} = 1 \qquad\Longrightarrow\qquad \frac{dy}{dx} = \frac{1}{f'(y)}$$

The derivative of an inverse is the reciprocal of the original’s derivative, evaluated at the corresponding point.

That is where the inverse trigonometric derivatives come from:

For $\arcsin$: if $y = \arcsin x$ then $\sin y = x$, so $\cos y \tfrac{dy}{dx} = 1$, giving $\tfrac{dy}{dx} = \tfrac{1}{\cos y}$. Then $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$, and the standard result drops out.

The third line is the same trick on $e^y = x$, which is why $\ln$ differentiates to $\tfrac1x$ — a result that looks like it should need its own proof and does not.