JEE Mains Maths · Differentiation

Functional Equations and Derivative Constants

Finding a function from a relation it satisfies, and handling polynomials whose own derivative values appear as their coefficients.

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Why this matters

Fifteen PYQs, nine of them multiple choice, and two from 2026. Eight start from a relation the function satisfies — in x and y, in x and 1/x, or in x and x + 1 — and find the function or its derivative; seven are polynomials, four of which carry their own derivative values as coefficients while three match coefficients or evaluate f and f′ directly. Two ideas cover the page.

Concept 1 of 2: Functional equations

An equation like

f(x+y)=f(x)f(y)

holds for every

x

and

y

, so choose convenient values. Put

x=y=0

to find

f(0)

. Then either recognise the standard solution, or write

f(x+h)

with the equation and take the limit that defines

f'(x)

: that gives a differential equation for

f

. When

f(x)

appears with

f\left(\frac1x\right)

, replace

x

\frac1x

and solve the two equations together.

Definition

$f(x+y)=f(x)+f(y)$ , differentiable: $f(x)=kx$ .
$f(x+y)=f(x)f(y)$ , never zero: $f(x)=e^{kx}$ with $k=f'(0)$ .
$f(xy)=f(x)+f(y)$ for $x>0$ : $f(x)=k\ln x$ .
$f(x)$ with $f\left(\frac1x\right)$ : swap $x$ and $\frac1x$ , then solve two linear equations.

Derivative from the equation

f(x+y)=f(x)f(y)\ \Rightarrow\ f'(x)=\lim_{h\to0}f(x)\frac{f(h)-1}{h}=f'(0)\,f(x)

Worked example

f(x+y)=f(x)+f(y)+xy

for all

x,y

, and

f'(0)=1

. Find

f(x)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 22 Jan 2025 · Q66Moderate

Example 1 · Differentiation · Functional Equations and Derivative Constants

Let

f(x)

be a real differentiable function such that

f(0) = 1

and

f(x + y) = f(x)f^{'}(y) +f^{'}(x)f(y)

for all

x,y \in R

. Then

\sum_{n = 1}^{100} \log_{e}f(n)

is equal to :

Find f(0) first

Most of these equations fix

f(0)

before anything else:

f(0)=f(0)^2

gives

f(0)=1

when

f

never vanishes, and

f(0)=2f(0)

gives

f(0)=0

. Skipping it leaves an unknown constant in the answer.

Drill 7 more on functional equations

Concept 2 of 2: Derivative values as constants

f(x)=x^3+x^2f'(1)+xf''(2)+f'''(3)

, the values

f'(1)

f''(2)

and

f'''(3)

are just numbers. Name them

a

b

c

, differentiate the cubic, and evaluate at 1, 2 and 3 to get equations for them. Start from the top:

f'''

of a cubic is a constant, so

c

comes first.

Definition

Replace each derivative value by a letter: it is a constant.
Differentiate the polynomial, then evaluate at the stated points.
Solve from the highest derivative down.
Given values of $f$ , $f'$ , $f''$ at one point fix the coefficients the same way.

For a cubic

f(x)=x^3+ax^2+bx+c\ \Rightarrow\ f'(x)=3x^2+2ax+b,\quad f''(x)=6x+2a,\quad f'''(x)=6

Worked example

f(x)=x^3+x^2f'(0)+xf''(1)+f'''(2)

. Find

f(x)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 2 · Q53Moderate

Example 2 · Differentiation · Functional Equations and Derivative Constants

Let

f(x) =x^{3}+x^{2}f^{'}(1) + 2xf^{''}(2) +f^{'''}(3)

x \in R

. Then the value of

f^{'}(5)

is :

The coefficient is not a function

f'(1)

is a number, so

x^2f'(1)

differentiates to

2xf'(1)

. Treating

f'(1)

as a function of

x

and using the product rule gives wrong equations.

Drill 6 more on derivative values as constants

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Functional equations
Derivative from the equation
$f(x+y)=f(x)f(y)\ \Rightarrow\ f'(x)=\lim_{h\to0}f(x)\frac{f(h)-1}{h}=f'(0)\,f(x)$
Derivative values as constants
For a cubic
$f(x)=x^3+ax^2+bx+c\ \Rightarrow\ f'(x)=3x^2+2ax+b,\quad f''(x)=6x+2a,\quad f'''(x)=6$

Watch out for (2)

Find f(0) first
→ Functional equations
The coefficient is not a function
→ Derivative values as constants

Test yourself on Differentiation

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