expresses both coordinates through a parameter rather than one through the other.

Each situation calls for a technique that adapts the core rules to a nonstandard setting. Implicit differentiation applies the chain rule through an equation without isolating

y

. Logarithmic differentiation converts multiplicative complexity into additive simplicity. The inverse function derivative formula recovers the slope of

f^{-1}

from the slope of

f

. Parametric differentiation computes

dy/dx

when neither variable is expressed directly in terms of the other.

Key Terms

Derivative— the target quantity these techniques compute

Differentiability— implicit differentiation requires the implicit function to be differentiable

See All Calculus Definitions →

Implicit Differentiation

Given an equation relating

x

and

y

—such as

x^2 + y^2 = 25

—implicit differentiation finds

\frac{dy}{dx}

without solving for

y

.

The method: differentiate both sides of the equation with respect to

x

. Every term involving only

x

is differentiated normally. Every term involving

y

is differentiated using the chain rule, treating

y

as a function of

x

. This introduces a factor of

\frac{dy}{dx}

wherever

y

appears. After differentiation, solve the resulting equation algebraically for

\frac{dy}{dx}

.

For

x^2 + y^2 = 25

: differentiating gives

2x + 2y\frac{dy}{dx} = 0

. Solving yields

\frac{dy}{dx} = -\frac{x}{y}

. The result depends on both

x

and

y

—this is typical and expected. The formula gives the slope at any point

(x, y)

on the curve without choosing a branch.

The technique works because the equation defines

y

as a function of

x

locally (by the Implicit Function Theorem), even when no global explicit formula exists. Differentiating the equation preserves the relationship while extracting the rate of change. The table below collects this procedure as a four-step recipe.

Step	What you do	Result
1	differentiate both sides of the equation with respect to x	a new equation whose terms involve x, y, and dy ⁄ dx
2	apply the chain rule to every term involving y, treating y as a function of x	each y-term picks up a factor of dy ⁄ dx
3	collect every dy ⁄ dx term on one side of the equation	dy ⁄ dx appears as a common factor on one side
4	solve algebraically for dy ⁄ dx	an expression for the slope at any point (x, y) on the curve

Notation in the Techniques

Notation

Notation in the Techniques

Three marks that make or break these methods: the hidden

y(x)

dependence, the parenthesised inverse-prime, and the parametric ratio with its false second-derivative twin.

\frac{dy}{dx}

and its quotient reading — differentials;

f^{-1}

itself — inverse functions;

\dot{y}

and the operator

\frac{d}{dx}

— higher-order derivatives.

2y\,\frac{dy}{dx}

Two y, times d y d x

The implicit convention: inside an equation being differentiated, every

y

silently means

y(x)

. So

\frac{d}{dx}[y^2] = 2y\,\frac{dy}{dx}

— the chain-rule factor appears wherever

y

does, as in Implicit Differentiation above.

CasesIn related rates the hidden variable is time: every quantity carries

\frac{d\cdot}{dt}

instead. In logarithmic differentiation the same convention produces

\frac{1}{y}\frac{dy}{dx}

from

\ln y

Also written

2y\,y'

— prime shorthand, compact in long computations but silent about the variable of differentiation.

Do not confuseTreating

y

as an independent symbol:

\frac{d}{dx}[y^2] = 2y

, dropping the factor. The single most common implicit-differentiation error — the missing

\frac{dy}{dx}

cannot be recovered later.

(f^{-1})'(b)

The derivative of f inverse, at b

Parentheses first, prime second: invert the function, then differentiate. The bracketing is load-bearing — three different objects live one keystroke apart.

Cases

(f^{-1})'

— derivative of the inverse, this page.

(f')^{-1}

— could be reciprocal or the inverse function of

f'

; context must decide.

\frac{1}{f'}

— the pointwise reciprocal, unambiguous.

Also written

\frac{dx}{dy}

— Leibniz form, where inverting the function literally flips the fraction:

\frac{dx}{dy} = 1\big/\frac{dy}{dx}

Do not confuseReading

(f^{-1})'(b) = \frac{1}{f'(a)}

as a pointwise identity at one input. The two sides are evaluated at different points —

b

on the left,

a = f^{-1}(b)

on the right; the

-1

superscript hazard itself is mapped on inverse functions.

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

d y d x equals d y d t over d x d t

The parametric slope: differentiate each coordinate with respect to the parameter, then divide. A literal fraction manipulation — the cancellation of

dt

is licensed by differentials.

CasesValid while

\frac{dx}{dt} \neq 0

; where it vanishes the tangent turns vertical. Physics writes the same ratio with Newton dots:

\frac{\dot{y}}{\dot{x}}

Also written

\frac{y'(t)}{x'(t)}

— prime-with-argument, safe because the argument names the variable.

Do not confuseThe false twin

\frac{d^2y}{dx^2} = \frac{d^2y/dt^2}{d^2x/dt^2}

. Second derivatives do not divide — the correct chain is in Parametric Differentiation below, and second-order differentials explain why the cancellation stops working past first order.

All calculus symbols →↑ Derivatives

Applications of Implicit Differentiation

Implicit differentiation extends beyond circles and ellipses. Any equation relating

x

and

y

that defines a smooth curve can be differentiated implicitly.

Tangent lines to implicitly defined curves follow directly. For

x^3 + y^3 = 6xy

(the folium of Descartes), implicit differentiation gives

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

. Evaluating at a specific point on the curve produces the tangent slope there.

Higher-order derivatives can be found implicitly as well. After finding

\frac{dy}{dx}

, differentiate the result implicitly again with respect to

x

—every occurrence of

\frac{dy}{dx}

is itself a function of

x

, and the first-derivative expression can be substituted back in. The algebra is heavier, but the method is systematic.

Related rates problems are implicit differentiation with respect to time

t

rather than

x

. If a relationship holds among several quantities that all vary with

t

, differentiating implicitly with respect to

t

connects their rates of change. This application is developed in graph analysis.

Logarithmic Differentiation

Logarithmic differentiation uses the properties of

\ln

to simplify differentiation of complex products, quotients, and variable-exponent expressions.

The procedure: given

y = f(x)

, take

\ln

of both sides to get

\ln y = \ln f(x)

. Apply logarithm properties—products become sums, quotients become differences, exponents become multipliers. Differentiate both sides implicitly with respect to

x

. The left side gives

\frac{1}{y}\frac{dy}{dx}

. Solve for

\frac{dy}{dx} = y \cdot \frac{d}{dx}[\ln f(x)]

Logarithmic Differentiation

y = f(x) \implies \ln y = \ln f(x) \implies \frac{y'}{y} = \frac{d}{dx}[\ln f(x)] \implies y' = y \cdot \frac{d}{dx}[\ln f(x)]

Learn more about this formula: Logarithmic Differentiation →

Browse all derivative formulas →

For

y = x^x

: taking

\ln

gives

\ln y = x \ln x

. Differentiating:

\frac{1}{y}\frac{dy}{dx} = \ln x + 1

. Solving:

\frac{dy}{dx} = x^x(\ln x + 1)

. No standard rule handles

x^x

directly—the base and exponent both vary. Logarithmic differentiation is the natural approach.

The technique also simplifies expressions like

y = \frac{x^2 \sqrt{x+1}}{(x-3)^4}

, where the product and quotient rules together would produce unwieldy algebra. After taking

\ln

, the expression becomes

\ln y = 2\ln x + \frac{1}{2}\ln(x+1) - 4\ln(x-3)

, and differentiating this sum is straightforward.

Differentiating Inverse Functions

f

is a one-to-one differentiable function with inverse

f^{-1}

, the derivative of the inverse is

Inverse Function Derivative

(f^{-1})'(b) = \frac{1}{f'(a)} \quad \text{where } b = f(a)

Learn more about this formula: Inverse Function Derivative →

Browse all derivative formulas →

provided

f'(f^{-1}(x)) \neq 0

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

The derivation uses implicit differentiation. If

y = f^{-1}(x)

, then

f(y) = x

. Differentiating both sides with respect to

x

f'(y) \cdot \frac{dy}{dx} = 1

, so

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

.

Geometrically, the graphs of

f

and

f^{-1}

are reflections across the line

y = x

. If

f

has slope

m

at a point,

f^{-1}

has slope

1/m

at the reflected point. A horizontal tangent on

f

(slope

0

) corresponds to a vertical tangent on

f^{-1}

(slope undefined), which is why

f'(f^{-1}(x)) \neq 0

is required.

Deriving Inverse Trigonometric Derivatives

The inverse function formula, combined with implicit differentiation, produces the derivatives of all inverse trigonometric functions without memorizing separate formulas.

For

y = \arcsin x

: the defining equation is

\sin y = x

with

y \in [-\pi/2, \pi/2]

. Differentiating implicitly:

\cos y \cdot \frac{dy}{dx} = 1

, so

\frac{dy}{dx} = \frac{1}{\cos y}

. Since

\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}

(positive because

y

is in the first or fourth quadrant), the result is

\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1 - x^2}}

For

y = \arctan x

: the defining equation is

\tan y = x

. Differentiating:

\sec^2 y \cdot \frac{dy}{dx} = 1

, so

\frac{dy}{dx} = \frac{1}{\sec^2 y} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}

.

The same method applies to

\arccos

\text{arccot}

\text{arcsec}

, and

\text{arccsc}

. In each case, implicit differentiation and a Pythagorean identity convert the result into an algebraic expression in

x

Parametric Differentiation

A curve defined parametrically by

x = x(t)

and

y = y(t)

does not express

y

as a function of

x

directly. The slope of the curve at a point is obtained through the chain rule in Leibniz form:

Parametric First Derivative

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} \quad \text{when } x = x(t), \, y = y(t)

Learn more about this formula: Parametric First Derivative →

provided

\frac{dx}{dt} \neq 0

. Each coordinate is differentiated with respect to the parameter

t

, and the ratio gives the slope.

For a circle parametrized by

x = \cos t

y = \sin t

\frac{dx}{dt} = -\sin t

and

\frac{dy}{dt} = \cos t

, so

\frac{dy}{dx} = \frac{\cos t}{-\sin t} = -\cot t

. At

t = \pi/4

, the slope is

-1

.

The second derivative of a parametric curve is not

\frac{d^2y/dt^2}{d^2x/dt^2}

. The correct formula is

Parametric Second Derivative

\frac{d^2 y}{dx^2} = \frac{d}{dx}\!\left[\frac{dy}{dx}\right] = \frac{d/dt\,[dy/dx]}{dx/dt}

Learn more about this formula: Parametric Second Derivative →

Browse all derivative formulas →

Differentiate

dy/dx

(which is a function of

t

) with respect to

t

, then divide by

dx/dt

once more. This error is common and produces incorrect concavity analysis when made.

When to Use Which Technique

Each technique targets a specific structural pattern.

Implicit differentiation applies when

x

and

y

are tangled in an equation that is difficult or impossible to solve for

y

—circles, ellipses, higher-degree curves, and any relation not naturally in

y = f(x)

form. It also applies in related rates, where multiple quantities depend on time.

Logarithmic differentiation applies when the expression involves products of many factors, quotients with complex structure, or—most distinctively—variable exponents like

x^x

(\sin x)^{\cos x}

, or

x^{\ln x}

. If the exponent contains the variable, logarithmic differentiation is typically the only viable approach.

The inverse function derivative applies when differentiating

f^{-1}

and the derivative of

f

is known. It is the standard route to inverse trigonometric and inverse hyperbolic derivatives.

Parametric differentiation applies when a curve is given as

x = x(t)

y = y(t)

. It handles curves that loop, cross themselves, or cannot be written as

y = f(x)

—cycloids, epicycloids, Lissajous figures, and motion trajectories.

These techniques are not mutually exclusive. A single problem may require implicit differentiation inside a parametric setting, or logarithmic differentiation as part of an inverse function computation. The table below lays out each technique alongside the signal that triggers it, the structural pattern it handles, and a typical example.

Technique	Signal that triggers it	What it handles	Typical example
Implicit differentiation	x and y are tangled in an equation that is hard to solve for y	curves not in y = f(x) form; related-rates problems	x² + y² = 25
Logarithmic differentiation	variable in the exponent, or many factors in a product or quotient	variable-exponent forms; complex products and quotients	y = x^x, y = (sin x)^{cos x}
Inverse function derivative	need the derivative of f⁻¹ and f' is known	inverse trigonometric and inverse hyperbolic derivatives	(arcsin x)' = 1 ⁄ √(1 − x²)
Parametric differentiation	a curve is given as x = x(t), y = y(t)	curves that loop or self-cross; motion trajectories	x = cos t, y = sin t

All Four Techniques as the Chain Rule

Each technique on this page looks different, but all four are expressions of the same underlying rule — the chain rule applied through a layer that hides the direct y-as-function-of-x form. The table below names that layer for each technique and the chain-rule form it produces.

Technique	Chain rule manifestation	What gets chained
Implicit differentiation	d ⁄ dx [g(y)] = g'(y) · dy ⁄ dx	y is treated as an inner function of x; the dy ⁄ dx factor is the chain rule's outer-inner derivative
Logarithmic differentiation	d ⁄ dx [ln f(x)] = f'(x) ⁄ f(x)	ln is chained with f, then implicit differentiation is applied to ln y = ln f(x)
Inverse function derivative	from f(f⁻¹(x)) = x: f'(f⁻¹(x)) · (f⁻¹)'(x) = 1	the identity f ∘ f⁻¹ = x is differentiated via the chain rule and rearranged
Parametric differentiation	dy ⁄ dx = (dy ⁄ dt) · (dt ⁄ dx) = (dy ⁄ dt) ⁄ (dx ⁄ dt)	the parameter t sits between y and x as the inner variable; chain rule is read off in Leibniz form

Differentiation Techniques FAQ

Why does d/dx[y²] equal 2y·dy/dx and not 2y?

Because inside an equation being differentiated implicitly, every y silently means y(x) — a function of x, not an independent symbol. Differentiating y² therefore needs the chain rule: the outer power contributes 2y and the inner function contributes its own derivative dy/dx. Dropping that factor is the most common implicit-differentiation error, and the missing dy/dx cannot be recovered later.Read more →

What is the difference between (f⁻¹)′ and 1/f′?

The parentheses are load-bearing. (f⁻¹)′ means invert the function first, then differentiate; 1/f′ is the pointwise reciprocal of the derivative. They are related by (f⁻¹)′(b) = 1/f′(a), but the two sides are evaluated at different points — b on the left, a = f⁻¹(b) on the right. Reading it as one shared input is the usual mistake.Read more →

How do you find dy/dx for a parametric curve?

Differentiate each coordinate with respect to the parameter and divide: dy/dx = (dy/dt)/(dx/dt). The cancellation of dt is legitimate here, licensed by differentials. The formula holds wherever dx/dt ≠ 0; where dx/dt vanishes the tangent turns vertical. Physics writes the same ratio with Newton dots, as ẏ/ẋ.Read more →

How do you find the second derivative of a parametric curve?

Not by dividing d²y/dt² by d²x/dt² — second derivatives do not divide that way. Instead differentiate the first-derivative expression dy/dx with respect to t, then divide by dx/dt once more. The dt cancellation that licenses the first-derivative ratio stops working past first order, which is exactly why the naive formula fails.Read more →