How are functional equations of the form f(x) + f(1/x) = g(x) solved?

Substitute x to 1/x in the given relation to obtain a paired equation. Solve the two linear equations for f(x) explicitly, then differentiate.

What is meant by an identity in x?

An identity in x means the equation holds for every real x. Coefficients of like independent terms (such as cos x, sin x, or powers of x) must match on both sides, giving a system of equations for the unknowns.

Derivatives Of Functions: Some Important Forms

Q: When should I prefer parametric differentiation over eliminating the parameter?

Use parametric differentiation whenever eliminating t produces a messy algebraic expression. For example, x = a cos^3 t, y = a sin^3 t gives dy/dx = -tan t parametrically, which is far cleaner than eliminating t.

Q: What is the chain rule and why is it central to JEE calculus?

The chain rule states that if y = F(u) and u = phi(x), then dy/dx = (dy/du)(du/dx). Almost every JEE calculus problem involves compositions such as sin(2x), e^(x^2) or log(1+x^2), so mastering the chain rule is essential.

Q: How do I differentiate a function like y = x^x?

Use logarithmic differentiation. Take ln of both sides: ln y = x ln x. Differentiate: (1/y)(dy/dx) = ln x + 1. Multiply by y to get dy/dx = x^x (ln x + 1).

Q: When do I use implicit differentiation instead of explicit?

Use implicit differentiation when the given relation F(x, y) = 0 cannot be solved uniquely for y (like x^2 + y^2 = 25). For every y-term, multiply by dy/dx, then isolate.

Q: Are the six inverse trigonometric derivatives all needed for JEE?

Yes, all six are in the JEE Main and JEE Advanced syllabus. Memorise the three primary ones (sin inverse, tan inverse, sec inverse) and recall that the derivatives of the co-functions are simply their negatives.

Q: What is the physical meaning of the second derivative?

If x = f(t) is position with time, dx/dt is velocity and d^2x/dt^2 is acceleration. Geometrically, the second derivative measures concavity: positive means concave up, negative means concave down.

Q: How do I differentiate one function with respect to another?

To find dy/dt where y = f(x) and t = g(x), compute dy/dx and dt/dx separately, then divide: dy/dt equals (dy/dx) divided by (dt/dx).

MathsLimits And DerivativesFor JEE aspirants

When a function is given in parametric, implicit, composite, inverse or logarithmic form, the basic power and sum rules alone are not enough. This concept covers all the important techniques JEE Main and JEE Advanced aspirants need to handle every such form: chain rule, implicit differentiation, parametric derivatives, inverse trigonometric derivatives, higher-order derivatives, logarithmic differentiation for $u^{v}$ type expressions, and differentiation of one function with respect to another.

Key Formulas — Quick Reference

Parametric: if $x = f (t)$ , $y = g (t)$ , then $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ (when $\frac{d x}{d t} \neq = 0$ )
Chain rule: if $y = F (u)$ and $u = ϕ (x)$ , then $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$
Implicit: differentiate both sides of $F (x, y) = 0$ w.r.t. $x$ treating $y$ as function of $x$ , then solve for $\frac{d y}{d x}$
Inverse trig derivatives (each with its domain restriction):
   $\frac{d}{d x} (sin^{- 1} x) = \frac{1}{1 - x ^{2}}$ , $\frac{d}{d x} (cos^{- 1} x) = - \frac{1}{1 - x ^{2}}$
   $\frac{d}{d x} (tan^{- 1} x) = \frac{1}{1 + x ^{2}}$ , $\frac{d}{d x} (cot^{- 1} x) = - \frac{1}{1 + x ^{2}}$
   $\frac{d}{d x} (sec^{- 1} x) = \frac{1}{∣ x ∣ x ^{2} - 1}$ , $\frac{d}{d x} (csc^{- 1} x) = - \frac{1}{∣ x ∣ x ^{2} - 1}$
Higher order: $\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d x} (\frac{d y}{d x})$ , $\frac{d ^{n} y}{d x ^{n}} = y_{n}$
Logarithmic (for $y = u^{v}$ ): $ln y = v ln u$ , then differentiate both sides
Function w.r.t. function: $\frac{d y}{d t} = \frac{d y / d x}{d t / d x}$ where $y = f (x)$ and $t = g (x)$

1. Derivatives of Functions in Parametric Form

Sometimes $x$ and $y$ are both expressed as functions of a third variable $t$ (the parameter), such as $x = f (t)$ , $y = g (t)$ . Rather than eliminating $t$ and expressing $y$ directly in terms of $x$ (often algebraically messy), we can differentiate parametrically.

\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{g ^{'} ( t )}{f ^{'} ( t )}, \frac{d x}{d t} \neq = 0

This follows from the chain rule: $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ , so $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ .

Solved Example 1

Find

\frac{d y}{d x}

x = 2 a t^{2}

and

y = 2 a t^{3}

Solution:

Differentiating each with respect to $t$ :

\frac{d x}{d t} = 4 a t, \frac{d y}{d t} = 6 a t^{2}

Therefore

\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{6 a t ^{2}}{4 a t} = \frac{3 t}{2}

Solved Example 2

x = e^{- t^{2}}

and

y = tan^{- 1} (2 t + 1)

, find

\frac{d y}{d x}

Solution:

For $x = e^{- t^{2}}$ : applying the chain rule with inner function $- t^{2}$ ,

\frac{d x}{d t} = e^{- t^{2}} \cdot (- 2 t) = - 2 t e^{- t^{2}}

For $y = tan^{- 1} (2 t + 1)$ :

\frac{d y}{d t} = \frac{1}{1 + ( 2 t + 1 ) ^{2}} \cdot 2 = \frac{2}{1 + ( 2 t + 1 ) ^{2}} = \frac{2}{4 t ^{2} + 4 t + 2} = \frac{1}{2 t ^{2} + 2 t + 1}

Therefore

\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{1/ ( 2 t ^{2} + 2 t + 1 )}{- 2 t e ^{- t^{2}}} = - \frac{e ^{t^{2}}}{2 t ( 2 t ^{2} + 2 t + 1 )}

2. Derivative of a Composite Function (Chain Rule)

Given a composite function $y = F (u)$ where $u = ϕ (x)$ , the derivative is

\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} = F^{'} (u) \cdot ϕ^{'} (x)

The rule extends to any number of nested compositions. For $y = f (u (v (x)))$ :

\frac{d y}{d x} = \frac{df}{d u} \cdot \frac{d u}{d v} \cdot \frac{d v}{d x}

Figure 1: The chain rule for

y = F (ϕ (x))

multiplies the local rate

\frac{d y}{d u}

by the inner rate

\frac{d u}{d x}

Solved Example 3

Find

\frac{d y}{d x}

for:
(i)

y = sin cos x

(ii)

y = sin (lo g \frac{x}{x + 1})

(iii)

y = lo g (x + e^{x})

Solution:

(i) Layer by layer: outer $sin$ , then , then $cos x$ .

\frac{d y}{d x} = cos cos x \cdot \frac{1}{2 cos x} \cdot (- sin x) = - \frac{sin x \cdot cos cos x}{2 cos x}

(ii) Simplify first using $lo g u = \frac{1}{2} lo g u$ :

y = sin (\frac{1}{2} lo g \frac{x}{x + 1}) = sin (\frac{1}{2} [lo g x - lo g (x + 1)])

Now differentiate:

\frac{d y}{d x} = cos (\frac{1}{2} lo g \frac{x}{x + 1}) \cdot \frac{1}{2} (\frac{1}{x} - \frac{1}{x + 1}) = \frac{1}{2 x ( x + 1 )} cos (lo g \frac{x}{x + 1})

(iii) Outer $lo g$ , then inner $x + e^{x}$ :

\frac{d y}{d x} = \frac{1}{x + e ^{x}} [1 + e^{x} \cdot \frac{1}{2 x}] = \frac{2 x + e ^{x}}{2 x ( x + e ^{x} )}

3. Derivative of an Implicit Function

If $x$ and $y$ satisfy a relation $F (x, y) = 0$ from which $y$ cannot be extracted as a single expression in $x$ , then $y$ is called an implicit function of $x$ . For example, $x^{2} + y^{2} = r^{2}$ gives $y = \pm r^{2} - x^{2}$ - two values, so $y$ is not uniquely a function of $x$ .

To find $\frac{d y}{d x}$ : differentiate both sides of the equation with respect to $x$ , treating $y$ as a differentiable function of $x$ (so every $y$ -term uses the chain rule and picks up a $\frac{d y}{d x}$ factor). Then solve algebraically for $\frac{d y}{d x}$ .

Example illustration: for

x^{3} + y^{2} = a^{2}

, differentiate to get

3 x^{2} + 2 y \frac{d y}{d x} = 0

, so

\frac{d y}{d x} = - \frac{3 x ^{2}}{2 y}

Solved Example 4

Find

\frac{d y}{d x}

for:
(i)

lo g (x y) = x^{2} + y^{2}

(ii)

x + y = sin (x y)

Solution:

(i) Using $lo g (x y) = lo g x + lo g y$ , the equation becomes $lo g x + lo g y = x^{2} + y^{2}$ . Differentiating both sides:

\frac{1}{x} + \frac{1}{y} \frac{d y}{d x} = 2 x + 2 y \frac{d y}{d x}

Collect the $\frac{d y}{d x}$ terms:

(\frac{1}{y} - 2 y) \frac{d y}{d x} = 2 x - \frac{1}{x}

\frac{d y}{d x} = \frac{2 x - \frac{1}{x}}{\frac{1}{y} - 2 y} = \frac{y ( 2 x ^{2} - 1 )}{x ( 1 - 2 y ^{2} )}

(ii) Differentiating $x + y = sin (x y)$ with respect to $x$ :

1 + \frac{d y}{d x} = cos (x y) \cdot (x \frac{d y}{d x} + y)

\frac{d y}{d x} (1 - x cos (x y)) = y cos (x y) - 1

\frac{d y}{d x} = \frac{y cos ( x y ) - 1}{1 - x cos ( x y )}

Solved Example 5

x^{2} + y^{2} + x y = 2

, find

\frac{d y}{d x}

Solution:

Differentiating both sides with respect to $x$ :

2 x + 2 y \frac{d y}{d x} + (x \frac{d y}{d x} + y) = 0

Collect $\frac{d y}{d x}$ :

(2 y + x) \frac{d y}{d x} = - (2 x + y)

\frac{d y}{d x} = - \frac{2 x + y}{2 y + x}

4. Inverse Functions and Their Derivatives

Theorem. If

y = f (x)

and

x = g (y)

are mutual inverses, and

f^{'} (x)

exists with

f^{'} (x) \neq = 0

, then

g^{'} (y) = \frac{1}{f ^{'} ( x )}

, that is,

\frac{d x}{d y} = \frac{1}{d y / d x}

Applying this to the six inverse trigonometric functions gives the standard derivatives:

Function	Derivative	Domain
$sin^{- 1} x$	$\frac{1}{1 - x ^{2}}$	$- 1 < x < 1$
$cos^{- 1} x$	$- \frac{1}{1 - x ^{2}}$	$- 1 < x < 1$
$tan^{- 1} x$	$\frac{1}{1 + x ^{2}}$	$x \in R$
$cot^{- 1} x$	$- \frac{1}{1 + x ^{2}}$	$x \in R$
$sec^{- 1} x$	$\frac{1}{∣ x ∣ x ^{2} - 1}$	$∣ x ∣ > 1$
$csc^{- 1} x$	$- \frac{1}{∣ x ∣ x ^{2} - 1}$	$∣ x ∣ > 1$

Solved Example 6

Find

\frac{d y}{d x}

for each function:
(a)

y = sin 2 x

(b)

y = cos 2 x

(c)

y = x sin^{- 1} x

(d)

y = a tan^{- 1} x

(e)

y = sec^{2} x

(f)

y = (lo g_{2} x)^{2}

Solution:

(a) Chain rule: $\frac{d y}{d x} = cos 2 x \cdot 2 = 2 cos 2 x$

(b) $\frac{d y}{d x} = - sin 2 x \cdot 2 = - 2 sin 2 x$

(c) Product rule with $u = x$ , $v = sin^{- 1} x$ :

\frac{d y}{d x} = x \cdot \frac{1}{1 - x ^{2}} + sin^{- 1} x \cdot 1 = \frac{x}{1 - x ^{2}} + sin^{- 1} x

(d) $a$ is a constant multiplier: $\frac{d y}{d x} = \frac{a}{1 + x ^{2}}$

(e) Write $y = (sec x)^{2}$ ; using chain rule:

\frac{d y}{d x} = 2 sec x \cdot sec x tan x = 2 sec^{2} x tan x

(f) Chain rule with $u = lo g_{2} x$ :

\frac{d y}{d x} = 2 lo g_{2} x \cdot \frac{1}{x ln 2} = \frac{2 lo g _{2} x}{x ln 2}

5. Higher Order Derivatives

The derivative $\frac{d y}{d x}$ is itself a function of $x$ and can be differentiated again. Successive derivatives are written:

First derivative: $\frac{d y}{d x} = f^{'} (x) = y_{1}$

Second derivative: $\frac{d ^{2} y}{d x ^{2}} = f^{''} (x) = y_{2}$

Third derivative: $\frac{d ^{3} y}{d x ^{3}} = f^{'''} (x) = y_{3}$ , and so on.

Physical meaning: if

x = f (t)

is position as a function of time

t

, then

\frac{d x}{d t}

is velocity and

\frac{d ^{2} x}{d t ^{2}}

is acceleration. Higher-order derivatives are essential in kinematics, oscillations, and application-of-derivative problems.

Solved Example 7

Find the second derivative of

y = a x^{3} + b x^{2} + c x + d

Solution:

Differentiating term by term:

\frac{d y}{d x} = 3 a x^{2} + 2 b x + c

Differentiating again:

\frac{d ^{2} y}{d x ^{2}} = 6 a x + 2 b

6. Logarithmic Differentiation

When $u$ and $v$ are functions of $x$ , an expression of the form $y = u^{v}$ (variable base and variable exponent) cannot be handled by the power rule or the exponential rule alone. Take the natural logarithm of both sides, then differentiate.

For $y = u^{v}$ :

ln y = v ln u

Differentiate both sides with respect to $x$ :

\frac{1}{y} \frac{d y}{d x} = \frac{d v}{d x} ln u + v \cdot \frac{1}{u} \frac{d u}{d x}

Multiply by $y = u^{v}$ to isolate $\frac{d y}{d x}$ .

Log differentiation is also useful whenever a function is a large product or quotient - taking log converts the product into a sum, which is easier to differentiate.

Solved Example 8

Differentiate

y = (lo g x)^{t a n x}

with respect to

z = sin (m cos^{- 1} x)

Solution:

First find $\frac{d y}{d x}$ by log differentiation. Take $ln$ of both sides:

ln y = tan x \cdot ln (lo g x)

Differentiate:

\frac{1}{y} \frac{d y}{d x} = sec^{2} x \cdot ln (lo g x) + tan x \cdot \frac{1}{lo g x} \cdot \frac{1}{x}

\frac{d y}{d x} = (lo g x)^{t a n x} [sec^{2} x \cdot ln (lo g x) + \frac{tan x}{x lo g x}]

Next, $\frac{d z}{d x}$ using the chain rule on $z = sin (m cos^{- 1} x)$ :

\frac{d z}{d x} = cos (m cos^{- 1} x) \cdot m \cdot (- \frac{1}{1 - x ^{2}}) = - \frac{m cos ( m cos ^{- 1} x )}{1 - x ^{2}}

Finally:

\frac{d y}{d z} = \frac{d y / d x}{d z / d x} = - \frac{( lo g x ) ^{t a n x} [ sec ^{2} x \cdot ln ( lo g x ) + \frac{tan x}{x lo g x} ] 1 - x ^{2}}{m cos ( m cos ^{- 1} x )}

Solved Example 9

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

, find

\frac{d y}{d x}

Solution:

Take $ln$ of both sides:

y ln x + x ln y = ln 1 = 0

Differentiate with respect to $x$ (treating $y$ as function of $x$ ):

y \cdot \frac{1}{x} + ln x \cdot \frac{d y}{d x} + ln y + x \cdot \frac{1}{y} \frac{d y}{d x} = 0

Collect $\frac{d y}{d x}$ :

(ln x + \frac{x}{y}) \frac{d y}{d x} = - (\frac{y}{x} + ln y)

\frac{d y}{d x} = - \frac{\frac{y}{x} + ln y}{ln x + \frac{x}{y}} = - \frac{y ( y + x ln y )}{x ( x + y ln x )}

7. Differentiating One Function with Respect to Another

To differentiate $y = f (x)$ with respect to another function $t = g (x)$ , use the identity

\frac{d y}{d t} = \frac{d y / d x}{d t / d x}

Compute $\frac{d y}{d x}$ and $\frac{d t}{d x}$ separately, then divide.

Solved Example 10

Differentiate

sin^{2} x

with respect to

(lo g x)^{2}

Solution:

Let $y = sin^{2} x$ and $t = (lo g x)^{2}$ .

\frac{d y}{d x} = 2 sin x cos x = sin 2 x

\frac{d t}{d x} = 2 lo g x \cdot \frac{1}{x} = \frac{2 lo g x}{x}

Therefore

\frac{d y}{d t} = \frac{d y / d x}{d t / d x} = \frac{sin 2 x}{\frac{2 lo g x}{x}} = \frac{x sin 2 x}{2 lo g x}

8. Tricky Forms and Functional Equations

Infinite nested radicals and continued expressions

When a function is defined by an infinite self-similar expression, note that the whole expression can be substituted for its inner copy - this converts an infinite tower into a finite equation.

Solved Example 11

y = sin x + sin x + sin x + \dots \infty

, find

\frac{d y}{d x}

Solution:

The inner nested expression is again $y$ , so

y = sin x + y \Rightarrow y^{2} = sin x + y

Differentiating both sides with respect to $x$ :

2 y \frac{d y}{d x} = cos x + \frac{d y}{d x}

(2 y - 1) \frac{d y}{d x} = cos x

\frac{d y}{d x} = \frac{cos x}{2 y - 1}

Functional equations

When $f (x)$ satisfies a relation such as $a f (x) + b f (\frac{1}{x}) = h (x)$ , substitute $x \to \frac{1}{x}$ to get a second equation, and solve the two linear equations for $f (x)$ .

Solved Example 12

5 f (x) + 3 f (\frac{1}{x}) = x + 2

and

y = x f (x)

, find

\frac{d y}{d x}

x = 1

Solution:

Given: $5 f (x) + 3 f (\frac{1}{x}) = x + 2$ …(i)

Replace $x$ by $\frac{1}{x}$ : $5 f (\frac{1}{x}) + 3 f (x) = \frac{1}{x} + 2$ …(ii)

Multiply (i) by $5$ and (ii) by $3$ , then subtract:

25 f (x) + 15 f (\frac{1}{x}) - 9 f (x) - 15 f (\frac{1}{x}) = 5 (x + 2) - 3 (\frac{1}{x} + 2)

16 f (x) = 5 x - \frac{3}{x} + 4 \Rightarrow f (x) = \frac{1}{16} (5 x - \frac{3}{x} + 4)

Then

y = x f (x) = \frac{1}{16} (5 x^{2} - 3 + 4 x)

\frac{d y}{d x} = \frac{1}{16} (10 x + 4)

At $x = 1$ : $\frac{d y}{d x} = \frac{14}{16} = \frac{7}{8}$ .

Identities in $x$

When an equation is stated as an identity in $x$ , it must hold for every value of $x$ . Match coefficients of like terms ( $cos x$ , $sin x$ , powers of $x$ ) on both sides.

Solved Example 13

Find

a, b, c, d

f (x) = (a x + b) cos x + (c x + d) sin x

and

f^{'} (x) = x cos x

is an identity in

x

Solution:

Differentiate $f (x)$ using product rule on each term:

f^{'} (x) = a cos x - (a x + b) sin x + c sin x + (c x + d) cos x

Group $cos x$ and $sin x$ terms:

f^{'} (x) = (a + c x + d) cos x + (c - a x - b) sin x

Setting this equal to the identity $x cos x + 0 \cdot sin x$ :

Coefficient of $cos x$ : $a + d + c x = x$ $\Rightarrow$ $c = 1$ and $a + d = 0$ .

Coefficient of $sin x$ : $c - b - a x = 0$ $\Rightarrow$ $a = 0$ and $c - b = 0$ $\Rightarrow$ $b = c = 1$ .

From $a + d = 0$ with $a = 0$ : $d = 0$ .

Therefore $a = 0, b = 1, c = 1, d = 0$ .

Common Mistakes to Avoid

Watch out

Chain rule missed on inner functions: $\frac{d}{d x} (sin 2 x) = 2 cos 2 x$ , not $cos 2 x$ . Always multiply by the derivative of the inner expression.
Implicit differentiation and the $\frac{d y}{d x}$ factor: when differentiating $y^{2}$ , the answer is $2 y \frac{d y}{d x}$ , not $2 y$ . Every $y$ -term picks up a chain-rule factor because $y$ is a function of $x$ .
Parametric derivative direction: $\frac{d y}{d x}$ equals $\frac{d y / d t}{d x / d t}$ , not $\frac{d x / d t}{d y / d t}$ . Check the denominator carefully.
Inverse-trig domain conditions: the formulas for $sec^{- 1}$ and $csc^{- 1}$ include $∣ x ∣$ , and the domain is $∣ x ∣ > 1$ . Ignoring these can lead to sign errors near boundary points.
Log differentiation without checking sign of $u$ : $ln u$ requires $u > 0$ . When $u$ can be negative, use $ln ∣ u ∣$ and be careful about the sign.
Confusing $u^{v}$ with $u^{n}$ or $a^{v}$ : $\frac{d}{d x} (x^{n})$ uses the power rule; $\frac{d}{d x} (a^{x})$ uses the exponential rule; $\frac{d}{d x} (u^{v})$ with both base and exponent variable requires log differentiation. Identify which case applies before choosing a method.
Higher derivative notation: $\frac{d ^{2} y}{d x ^{2}}$ is not $(\frac{d y}{d x})^{2}$ . It is the derivative of the derivative.
Second equation missed in functional problems: when a functional equation involves $f (x)$ and $f (\frac{1}{x})$ (or $f (1 - x)$ ), remember to derive the paired equation by substitution before solving.

Frequently Asked Questions

Q1. When should I prefer parametric differentiation over eliminating the parameter?

Use parametric differentiation whenever eliminating $t$ produces a messy algebraic expression. For instance, $x = a cos^{3} t$ , $y = a sin^{3} t$ (an astroid) gives a clean $\frac{d y}{d x} = - tan t$ parametrically, but eliminating $t$ leads to $x^{2/3} + y^{2/3} = a^{2/3}$ , which is harder to differentiate. Keep the parameter unless the problem explicitly demands otherwise.

Q2. What is the chain rule and why is it central to JEE calculus?

The chain rule gives the derivative of a composite function: if $y = F (u)$ and $u = ϕ (x)$ , then $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ . Almost every JEE calculus problem involves compositions ( $sin (2 x)$ , $e^{x^{2}}$ , $lo g (1 + x^{2})$ ), so mastering the chain rule is non-negotiable for both Main and Advanced.

Q3. How do I differentiate a function like $y = x^{x}$ ?

Use logarithmic differentiation. Take $ln$ of both sides: $ln y = x ln x$ . Differentiate: $\frac{1}{y} \frac{d y}{d x} = ln x + 1$ . Multiply by $y$ : $\frac{d y}{d x} = x^{x} (ln x + 1)$ . This method works for every $y = u^{v}$ with variable base and variable exponent.

Q4. When do I use implicit differentiation instead of explicit?

Use implicit differentiation when the given relation $F (x, y) = 0$ cannot be solved uniquely for $y$ (like $x^{2} + y^{2} = 25$ or $sin (x y) = x - y$ ), or when solving would be messy. For every $y$ -term you differentiate, multiply by $\frac{d y}{d x}$ , then isolate.

Q5. Are the six inverse trigonometric derivatives all needed for JEE?

Yes, all six are in the JEE Main and Advanced syllabus. Memorise the three primary ones ( $sin^{- 1}$ , $tan^{- 1}$ , $sec^{- 1}$ ) and remember that the derivatives of $cos^{- 1}$ , $cot^{- 1}$ , $csc^{- 1}$ are simply the negatives.

Q6. What is the physical meaning of the second derivative?

If position $x = f (t)$ depends on time $t$ , then $\frac{d x}{d t}$ is velocity and $\frac{d ^{2} x}{d t ^{2}}$ is acceleration. In geometry, the second derivative $f^{''} (x)$ measures the concavity of the curve: $f^{''} (x) > 0$ means concave up, $f^{''} (x) < 0$ means concave down. The sign is used in the second-derivative test for maxima and minima.

Q7. How do I differentiate one function with respect to another?

To find $\frac{d y}{d t}$ where $y = f (x)$ and $t = g (x)$ , compute $\frac{d y}{d x}$ and $\frac{d t}{d x}$ separately and then divide: $\frac{d y}{d t} = \frac{d y / d x}{d t / d x}$ . This is a direct consequence of the chain rule and is very common in JEE Advanced problems.

Q8. How do functional equations like $f (x) + f (1/ x) = g (x)$ get solved?

Substitute $x \to \frac{1}{x}$ (or $x \to 1 - x$ , whichever preserves the argument set) into the given relation. This produces a second linear equation in $f (x)$ and $f (1/ x)$ . Solve the pair as a $2 \times 2$ system for $f (x)$ explicitly, and then differentiate.

Q9. What is meant by "identity in $x$ " in the last example?

An identity in $x$ means the equation holds for every real $x$ , not just for a particular value. Under this condition, coefficients of like linearly-independent terms (such as $cos x$ , $sin x$ , or powers $1, x, x^{2}, \dots$ ) must match on both sides. Comparing coefficients gives a system for the unknowns.

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1. Derivatives of Functions in Parametric Form

2. Derivative of a Composite Function (Chain Rule)

3. Derivative of an Implicit Function

4. Inverse Functions and Their Derivatives

5. Higher Order Derivatives

6. Logarithmic Differentiation

7. Differentiating One Function with Respect to Another

8. Tricky Forms and Functional Equations

Infinite nested radicals and continued expressions

Functional equations

Identities in x

Common Mistakes to Avoid

Frequently Asked Questions

Q1. When should I prefer parametric differentiation over eliminating the parameter?

Q2. What is the chain rule and why is it central to JEE calculus?

Q3. How do I differentiate a function like y=xx?

Q4. When do I use implicit differentiation instead of explicit?

Q5. Are the six inverse trigonometric derivatives all needed for JEE?

Q6. What is the physical meaning of the second derivative?

Q7. How do I differentiate one function with respect to another?

Q8. How do functional equations like f(x)+f(1/x)=g(x) get solved?

Q9. What is meant by "identity in x" in the last example?

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Identities in $x$

Q3. How do I differentiate a function like $y = x^{x}$ ?

Q8. How do functional equations like $f (x) + f (1/ x) = g (x)$ get solved?

Q9. What is meant by "identity in $x$ " in the last example?