Implicit Differentiation

Not every relationship arrives as $y = f(x)$. In $x - y - \pi = 0$ we could solve for $y$; in $x + \sin(xy) - y = 0$ we cannot — yet $y$ still depends on $x$ implicitly. Implicit differentiation extracts $\frac{dy}{dx}$ without ever solving:

1. Differentiate both sides with respect to $x$, remembering that $y$ is a function of $x$ — so every appearance of $y$ triggers the chain rule and produces a $\frac{dy}{dx}$ factor. E.g. $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ and $\frac{d}{dx}(\sin y) = \cos y \frac{dy}{dx}$.
2. Collect the $\frac{dy}{dx}$ terms on one side.
3. Solve — the answer may legitimately contain both $x$ and $y$.

Worked instantly: $y + \sin y = \cos x$ gives $\frac{dy}{dx} + \cos y \frac{dy}{dx} = -\sin x$, so $$\frac{dy}{dx} = -\frac{\sin x}{1 + \cos y}, \qquad y \neq (2n+1)\pi$$

Derivatives of Inverse Trigonometric Functions

The same idea — differentiate the defining relation — produces the inverse-trig table. For $y = \sin^{-1} x$: write $x = \sin y$ and differentiate: $1 = \cos y \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - x^2}}$ (cosine is positive on the principal branch).

Note the pairing: $\sin^{-1}$ and $\cos^{-1}$ have opposite derivatives — consistent with $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$, whose derivative must be $0$.

Simplify first — the substitution reflex

When the argument of an inverse-trig function looks like a double-angle formula, substitute before differentiating:

1. $\sin^{-1}\left(\frac{2x}{1 + x^2}\right)$: put $x = \tan\theta$; the argument becomes $\sin 2\theta$, so the function is $2\tan^{-1} x$ (for $\vert x \vert < 1$) — derivative $\frac{2}{1 + x^2}$ in one line.
2. $\tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) = 3\tan^{-1} x$ (for $\vert x \vert < \frac{1}{\sqrt{3}}$) — derivative $\frac{3}{1 + x^2}$.
3. $\cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) = 2\tan^{-1} x$ (for $x > 0$) — derivative $\frac{2}{1 + x^2}$.

Differentiating these by brute quotient-and-chain is a page of algebra; the substitution makes each a one-liner. Watch the stated interval — the clean identity holds only on the branch the problem specifies.

Solved Examples

Example 1 — Implicit warm-up

Find $\dfrac{dy}{dx}$ if $2x + 3y = \sin x$.

Step 1 — differentiate both sides: $2 + 3\frac{dy}{dx} = \cos x$.

Step 2 — solve: $\frac{dy}{dx} = \dfrac{\cos x - 2}{3}$.

Answer: as above — the simplest implicit pattern: $y$ appears linearly, so one rearrangement finishes it.

Example 2 — Chain rule inside the implicit step

Find $\dfrac{dy}{dx}$ if $y + \sin y = \cos x$.

Step 1 — differentiate, chain rule on $\sin y$: $\frac{dy}{dx} + \cos y \frac{dy}{dx} = -\sin x$.

Step 2 — factor and solve: $(1 + \cos y)\frac{dy}{dx} = -\sin x$, so $$\frac{dy}{dx} = -\frac{\sin x}{1 + \cos y}, \qquad y \neq (2n+1)\pi$$

Answer: as displayed — the exclusion keeps the denominator nonzero, and the answer properly involves both variables.

Example 3 — A symmetric polynomial relation

Find $\dfrac{dy}{dx}$ if $x^2 + xy + y^2 = 100$.

Step 1 — differentiate (product rule on $xy$): $2x + \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0$.

Step 2 — collect: $(x + 2y)\frac{dy}{dx} = -(2x + y)$.

Answer: $\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}$ — the $xy$ term feeds both sides of the collection, one piece with $\frac{dy}{dx}$ and one without.

Example 4 — Deriving the $\sin^{-1}$ formula

Find the derivative of $f(x) = \sin^{-1} x$, assuming it exists.

Step 1 — invert the relation: $y = \sin^{-1} x$ means $x = \sin y$ with $y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Step 2 — differentiate implicitly: $1 = \cos y \frac{dy}{dx}$.

Step 3 — return to $x$: on the principal branch $\cos y \geq 0$, so $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$.

Answer: $\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^2}}$, valid for $-1 < x < 1$ — the endpoints are excluded because $\cos y = 0$ there.

Example 5 — The substitution reflex in action

Differentiate $y = \sin^{-1}\left(\dfrac{2x}{1 + x^2}\right)$ for $\vert x \vert < 1$.

Step 1 — substitute $x = \tan\theta$: then $\frac{2x}{1 + x^2} = \frac{2\tan\theta}{1 + \tan^2\theta} = \sin 2\theta$.

Step 2 — collapse: $y = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1} x$ (the interval $\vert x \vert < 1$ keeps $2\theta$ inside the principal branch).

Step 3 — differentiate the simplified form: $\frac{dy}{dx} = \frac{2}{1 + x^2}$.

Answer: $\dfrac{2}{1 + x^2}$ — three lines, against a page of quotient-rule algebra without the substitution.

Example 6 — The triple-angle cousin

Differentiate $y = \tan^{-1}\left(\dfrac{3x - x^3}{1 - 3x^2}\right)$ for $\vert x \vert < \dfrac{1}{\sqrt{3}}$.

Step 1 — recognise the pattern: with $x = \tan\theta$, the argument is exactly $\tan 3\theta$.

Step 2 — collapse: $y = 3\theta = 3\tan^{-1} x$ on the given interval.

Step 3 — differentiate: $\frac{dy}{dx} = \frac{3}{1 + x^2}$.

Answer: $\dfrac{3}{1 + x^2}$. The whole family — double angle, triple angle, $\frac{1 - x^2}{1 + x^2}$ — yields to the same three-line script.