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 with

y \neq 0

, without choosing a branch.

Pitfall: Dropping the dy/dx factor

Differentiating

y^2

2y

turns

x^2 + y^2 = 25

into

2x + 2y = 0

, an equation with no

\frac{dy}{dx}

to solve for. With

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

, the slope is

-\frac{x}{y}

, which is

-\frac{3}{4}

(3, 4)

The technique works because the equation defines

y

as a function of

x

locally (by the Implicit Function Theorem, near points where

\frac{\partial F}{\partial y} \neq 0

for the equation written as

F(x, y) = 0

), 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.

A slope from the equation alone

The section’s circle x² + y² = 25 is not the graph of one function: each x inside (−5, 5) has two points. Implicit differentiation gives dy/dx = −x/y, which uses both coordinates and so picks the right branch automatically. At (3, 4) the slope is −3/4, the dashed tangent line; at (3, −4) the same formula gives +3/4.

At (5, 0) the formula divides by y = 0: the tangent there is vertical.

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

The slope of a curve comes straight from its equation; there is no need to solve for

y

first.

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.

Pitfall: Forgetting the product rule on xy

Differentiating

6xy

6x\frac{dy}{dx}

gives

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

, a slope of

-3

at the point

(3, 3)

of the folium. With

\frac{d}{dx}[6xy] = 6y + 6x\frac{dy}{dx}

, the slope there is

-1

In the folium, the right side

6xy

is a product of two functions of

x

: the factor

x

itself and the factor

y

, which depends on

x

through the curve. The left side is differentiated term by term,

x^3

3x^2

and

y^3

3y^2\frac{dy}{dx}

. Once both sides are differentiated, every

\frac{dy}{dx}

term is collected on one side, and the slope formula above follows.

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 →

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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.

Pitfall: Forgetting to multiply by y

Stopping at

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

gives

\ln x + 1

as the derivative of

x^x

, about

1.69

x = 2

. Multiplying by

y = x^x

gives

x^x(\ln x + 1) \approx 6.77

there.

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.

Logarithms turn products, quotients and powers into sums, and sums are the easiest thing to differentiate.

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 →

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provided

f'(a) \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)}

Pitfall: Evaluating the derivative at the wrong point

Using

\frac{1}{f'(b)}

for

f(x) = e^x

and

b = 2

gives

(f^{-1})'(2) = \frac{1}{e^2} \approx 0.135

. The formula needs

a = f^{-1}(2) = \ln 2

, and

(f^{-1})'(2) = \frac{1}{e^{\ln 2}} = \frac{1}{2}

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.

Reciprocal slopes at mirrored points

eˣ and ln x are reflections of each other in the dashed line y = x. At (ln 2, 2) the curve eˣ has slope e^(ln 2) = 2; the mirrored point on ln x is (2, ln 2), where the slope is 1/2. Reflecting across y = x swaps rise and run, so every slope m becomes 1/m: (f⁻¹)′(b) = 1/f′(a).

A flat spot on f, slope 0, would mirror to a vertical tangent on f⁻¹, which is why f′ ≠ 0 is required.

Fact: A slope for an inverse with no formula

The function

f(x) = x^5 + x

is one-to-one, but no formula built from arithmetic and roots gives its inverse. Still,

f(1) = 2

and

f'(1) = 6

, so

(f^{-1})'(2) = \frac{1}{6}

exactly.

The formula never asks for

f^{-1}

itself. Everything on its right side comes from

f

: an input

a

, its output

b = f(a)

, and the slope

f'(a)

there. To find the slope of the inverse at

b

, it is enough to know which input of

f

produces

b

and to differentiate

f

at that input. Finding that input is the only step that involves the inverse at all, and it can often be done by trying simple values of

a

until

f(a)

equals

b

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

Derivative of Arcsine

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

Learn more about this formula: Derivative of Arcsine →

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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

Pitfall: Flipping the parametric ratio

Dividing

\frac{dx}{dt}

\frac{dy}{dt}

gives

\frac{-\sin t}{\cos t} = -\tan t

, which is

-\frac{1}{\sqrt{3}} \approx -0.577

t = \frac{\pi}{6}

. The slope is

\frac{dy/dt}{dx/dt} = -\cot t = -\sqrt{3} \approx -1.732

In the circle example the two derivatives play different roles. The rate

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

measures how fast the point rises, and

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

how fast it moves sideways; the slope is the rise per unit of sideways motion. Near

t = \pi/2

, at the top of the circle,

\frac{dy}{dt}

is close to

0

and the tangent is almost flat.

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 →

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Differentiate

dy/dx

(which is a function of

t

) with respect to

t

, then divide by

dx/dt

once more. Skipping that second division is a common error and produces incorrect concavity analysis.

A parametric slope is a ratio of two rates in

t

, but second derivatives cannot be divided the same way.

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.

Fact: Two wrong rules add up to the right one

Treating

x^x

as a power gives

x \cdot x^{x-1} = x^x

; treating it as an exponential gives

x^x \ln x

. Neither is its derivative, yet their sum

x^x(\ln x + 1)

is: when base and exponent both vary, their two effects add.

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 the standard approach; rewriting

f^g

e^{g \ln f}

is the same idea in another form.

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

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Before you read

Beyond the basic rules

Four techniques extend differentiation to functions the basic rules handle badly. Each fits a recognisable kind of problem.

Differentiation techniques4 methods

Implicit

When

x

and

y

are tangled, as in a circle

Alsorelated rates

Logarithmic

Whenmany factors or quotients

Alsovariable exponents such as

x^x

Inverse function

Whendifferentiating

f^{-1}

with

f'

known

Leads tothe inverse trig derivatives

Parametric

Whencurves given as

x(t)

y(t)

Examplesloops, cycloids, trajectories

Takeaway

Recognise the setting

Tangled

x

and

y

call for implicit differentiation; many factors or variable exponents call for logarithms; inverses and parametric curves each have their own route.

When to use which differentiation technique·/calculus/derivatives/techniquesLearn Math Class

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

Common Pitfalls

Six mistakes cause most wrong answers with these techniques. Five of them are marked where they happen in the sections above; the sixth, the parametric second derivative, is the one the section on parametric differentiation warns against. Here each is worked through in full, with the wrong and the right working side by side.

Dropping the dy/dx factor

Every term with

y

is differentiated as if

y

were the variable, so the chain-rule factor

\dfrac{dy}{dx}

never appears.

Wrong working: differentiating

x^2 + y^2 = 25

gives

2x + 2y = 0

. There is no

\dfrac{dy}{dx}

left to solve for, and the equation even fails at the point

(3, 4)

of the circle:

6 + 8 \neq 0

.

Right working:

y

is a function of

x

, so

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

. Then

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

and

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

; at

(3, 4)

the slope is

-\dfrac{3}{4}

.

Why it happens: in ordinary differentiation the letter in the expression is the variable. Here

x

is the variable, and

y

only looks like one.

How to check: where one branch can be solved for

y

, compare. On the upper half of the circle

y = \sqrt{25 - x^2}

, and

\dfrac{dy}{dx} = -\dfrac{x}{\sqrt{25 - x^2}}

, which is

-\dfrac{3}{4}

x = 3

.

Forgetting the product rule on xy

A term that multiplies

x

and

y

, like

6xy

in the folium

x^3 + y^3 = 6xy

, is differentiated as a constant times

y

, as though

x

did not change.

Wrong working:

\dfrac{d}{dx}[6xy] = 6x\dfrac{dy}{dx}

. Then

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

and

\dfrac{dy}{dx} = \dfrac{x^2}{2x - y^2}

, which at

(3, 3)

gives

\dfrac{9}{6 - 9} = -3

.

Right working:

6xy

is a product of two functions of

x

, so

\dfrac{d}{dx}[6xy] = 6y + 6x\dfrac{dy}{dx}

. Then

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

and

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

; at

(3, 3)

the slope is

\dfrac{18 - 27}{27 - 18} = -1

.

Why it happens: the

y

in the term draws all the attention, because it is the part that needs the chain rule, and the

x

beside it is treated as a coefficient.

How to check: the folium does not change when

x

and

y

are swapped, so it is symmetric about the line

y = x

. At the point

(3, 3)

on that line, the tangent must therefore have slope

1

-1

; the value

-3

cannot be right.

Forgetting to multiply by y

After the logarithm is differentiated, the result is the derivative of

\ln y

, not of

y

. The last step, multiplying through by

y

and writing

y

in terms of

x

, is left out.

Wrong working: from

\ln y = x\ln x

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

, so

\dfrac{dy}{dx} = \ln x + 1

.

Right working: multiply both sides by

y

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

.

Why it happens: the differentiated right side

\ln x + 1

looks like a finished answer, and the

\dfrac{1}{y}

in front of

\dfrac{dy}{dx}

on the left is easy to lose.

How to check: compare with a difference quotient at one point. At

x = 2

\dfrac{2.01^{2.01} - 1.99^{1.99}}{0.02} \approx 6.77

, which matches

4(\ln 2 + 1) \approx 6.77

and not

\ln 2 + 1 \approx 1.69

.

Evaluating the derivative at the wrong point

In

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

, the derivative of

f

is evaluated at

b

, the input of

f^{-1}

, instead of at

a = f^{-1}(b)

.

Wrong working: for

f(x) = e^x

f'(x) = e^x

, so

(f^{-1})'(2) = \dfrac{1}{f'(2)} = \dfrac{1}{e^2} \approx 0.135

.

Right working: first find

a

with

f(a) = 2

, which is

a = \ln 2

. Then

(f^{-1})'(2) = \dfrac{1}{f'(\ln 2)} = \dfrac{1}{e^{\ln 2}} = \dfrac{1}{2}

.

Why it happens: the formula is remembered as one over the derivative, and the point on the left side is carried over without being converted.

How to check: when

f^{-1}

is known, differentiate it directly. Here

f^{-1}(x) = \ln x

and

(\ln x)' = \dfrac{1}{x}

, which is

\dfrac{1}{2}

x = 2

.

Flipping the parametric ratio

The slope of a parametric curve is computed as

\dfrac{dx/dt}{dy/dt}

, with the two parameter derivatives in the wrong order.

Wrong working: for

x = \cos t

y = \sin t

\dfrac{dx/dt}{dy/dt} = \dfrac{-\sin t}{\cos t} = -\tan t

; at

t = \dfrac{\pi}{6}

this is

-\dfrac{1}{\sqrt{3}} \approx -0.577

.

Right working:

\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{\cos t}{-\sin t} = -\cot t

; at

t = \dfrac{\pi}{6}

the slope is

-\sqrt{3} \approx -1.732

.

Why it happens: both orders look like a ratio of two derivatives. At

t = \dfrac{\pi}{4}

, the point used in the section, the two orders even agree, both giving

-1

, so a check there catches nothing.

How to check: the tangent to a circle is perpendicular to the radius. At

t = \dfrac{\pi}{6}

the point is

\left(\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right)

, the radius has slope

\dfrac{1/2}{\sqrt{3}/2} = \dfrac{1}{\sqrt{3}}

, and the perpendicular slope is

-\sqrt{3}

.

Dividing the second derivatives

The second derivative of a parametric curve is computed by dividing

\dfrac{d^2y}{dt^2}

\dfrac{d^2x}{dt^2}

, copying the pattern of the first derivative.

Wrong working: for

x = \cos t

y = \sin t

\dfrac{d^2y/dt^2}{d^2x/dt^2} = \dfrac{-\sin t}{-\cos t} = \tan t

, which is

1

t = \dfrac{\pi}{4}

.

Right working: differentiate

\dfrac{dy}{dx} = -\cot t

with respect to

t

, which gives

\csc^2 t

, then divide by

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

. So

\dfrac{d^2y}{dx^2} = -\dfrac{1}{\sin^3 t}

, which is

-2\sqrt{2} \approx -2.83

t = \dfrac{\pi}{4}

.

Why it happens: the first-derivative formula looks like a cancellation of

dt

, and the same cancellation is expected one order higher, where it no longer holds.

How to check: the sign of the second derivative gives the concavity. At

t = \dfrac{\pi}{4}

the point

\left(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right)

is on the upper half of the circle, which bends downward, so the second derivative must be negative;

\tan t = 1

is positive.

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 →