Why is the power rule true for all real exponents?

For integer n it follows from the binomial theorem. For real n, use logarithmic differentiation: y = x^n gives ln y = n ln x, and differentiating both sides yields y' = n x^(n-1).

What is the difference between f'(x) and dy/dx?

They mean the same thing when y = f(x). Lagrange's f'(x) treats the derivative as a new function; Leibniz's dy/dx emphasises the ratio of infinitesimal changes.

Introduction to Derivatives

Q: What does the derivative of a function represent geometrically?

The derivative f'(x0) equals the slope of the tangent line to the curve y = f(x) at the point (x0, f(x0)). Physically, if x is time and y is position, then f'(x) is instantaneous velocity.

Q: When does a derivative not exist?

A derivative fails to exist at sharp corners (like |x| at 0), vertical tangents (like x^(1/3) at 0), or points where the function is discontinuous. Differentiability implies continuity but not vice versa.

Q: When should I use the product rule versus expanding first?

If the product is easy to expand (like (x+2)(x+3)), expand and differentiate term by term. Use the product rule when expansion is messy, such as for x^2 sin x.

Q: Can the derivative be negative or zero?

Yes. A negative derivative means the function is decreasing; a zero derivative means the tangent is horizontal, which happens at maxima, minima and inflection points.

Q: How many standard derivatives should I memorise for JEE?

At least twelve: x^n, constant, e^x, a^x, ln x, log_a x, and the six trigonometric functions. Add the six inverse-trig derivatives for full coverage.

MathsLimits And DerivativesFor JEE aspirants

The derivative of a function $f (x)$ at a point measures the instantaneous rate of change of $f$ with respect to $x$ , and equals the slope of the tangent line at that point. Formally, $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ when this limit exists finitely. This concept is the gateway to calculus for JEE Main and JEE Advanced aspirants: it powers tangent-normal problems, maxima-minima, motion analysis, and every applied problem where a quantity changes with another.

Key Formulas — Quick Reference

First principle: $f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$
Power rule: $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ for all real $n$
Constant: $\frac{d}{d x} (c) = 0$
Sum / Difference: $\frac{d}{d x} (u \pm v) = \frac{d u}{d x} \pm \frac{d v}{d x}$
Constant multiple: $\frac{d}{d x} (k u) = k \frac{d u}{d x}$
Product rule: $\frac{d}{d x} (uv) = u \frac{d v}{d x} + v \frac{d u}{d x}$
Quotient rule: $\frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}$ , $v \neq = 0$

1. Derivative from the First Principle

Let $y = f (x)$ be a function of $x$ . The derivative of $f$ at a point $x$ is defined as the limit

\frac{d y}{d x} = f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}

provided this limit exists finitely. This method is called differentiation from first principle or the ab initio method.

Alternative notations for the derivative of $y = f (x)$ include $y^{'}$ , $f^{'} (x)$ , $D f (x)$ and $\frac{d y}{d x}$ . The value of the derivative at a specific point $x = c$ is written $f^{'} (c)$ or $\frac{d y}{d x}_{x = c}$ .

Geometric Meaning: Slope of the Tangent

Take two points $P (x, f (x))$ and $Q (x + h, f (x + h))$ on the curve $y = f (x)$ . The slope of the secant line $P Q$ is

m_{P Q} = \frac{f ( x + h ) - f ( x )}{h}

As $h \to 0$ , the point $Q$ slides toward $P$ , and the secant $P Q$ becomes the tangent line at $P$ . Its slope is exactly $f^{'} (x)$ .

Figure 1: As

h \to 0

, the secant through

(x, f (x))

and

(x + h, f (x + h))

rotates into the tangent at

(x, f (x))

. Its slope is

f^{'} (x)

Solved Example 1

Find the derivative of

f (x) = x^{3}

using the first principle.

Solution:

By definition,

f^{'} (x) = h \to 0 lim \frac{( x + h ) ^{3} - x ^{3}}{h}

Expanding $(x + h)^{3} = x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3}$ :

f^{'} (x) = h \to 0 lim \frac{3 x ^{2} h + 3 x h ^{2} + h ^{3}}{h} = h \to 0 lim (3 x^{2} + 3 x h + h^{2}) = 3 x^{2}

Therefore $\frac{d}{d x} (x^{3}) = 3 x^{2}$ .

Solved Example 2

Find the derivative of

f (x) = \frac{1}{x}

from first principle.

Solution:

f^{'} (x) = h \to 0 lim \frac{\frac{1}{x + h} - \frac{1}{x}}{h} = h \to 0 lim \frac{x - x + h}{h x x + h}

Rationalising the numerator by multiplying with $\frac{x + x + h}{x + x + h}$ :

= h \to 0 lim \frac{x - ( x + h )}{h x x + h ( x + x + h )} = h \to 0 lim \frac{- 1}{x x + h ( x + x + h )}

As $h \to 0$ :

f^{'} (x) = \frac{- 1}{x \cdot x \cdot 2 x} = - \frac{1}{2 x ^{3/2}}

2. Standard Derivatives of Important Functions

Rather than deriving every result from first principle each time, JEE aspirants should memorise this list of standard derivatives. Every rule below can be proved from the definition; here we simply record the results.

Function $y = f (x)$	Derivative $\frac{d y}{d x}$	Function $y = f (x)$	Derivative $\frac{d y}{d x}$
$x^{n}$ (any real $n$ )	$n x^{n - 1}$	$sin x$	$cos x$
constant $c$	$0$	$cos x$	$- sin x$
$e^{x}$	$e^{x}$	$tan x$	$sec^{2} x$
$a^{x}$ ( $a > 0$ )	$a^{x} ln a$	$cot x$	$- csc^{2} x$
$ln x$ ( $x > 0$ )	$\frac{1}{x}$	$sec x$	$sec x tan x$
$lo g_{a} x$	$\frac{1}{x ln a}$	$csc x$	$- csc x cot x$

Note: The power rule

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

works for all real exponents

n

(positive, negative, fractional, irrational). Test it on

x^{5/2}

x^{- 3}

x^{e}

- the rule applies identically.

Solved Example 3

Find the derivative of

x^{5/2}

Solution:

Using the power rule with $n = \frac{5}{2}$ :

\frac{d}{d x} (x^{5/2}) = \frac{5}{2} x^{5/2 - 1} = \frac{5}{2} x^{3/2}

Solved Example 4

Find the derivative of

(1 + x)^{1/2}

treating

(1 + x)

as the base for the power rule.

Solution:

Applying the power rule to $(1 + x)^{1/2}$ with respect to the base $(1 + x)$ (the derivative of $1 + x$ w.r.t. $x$ is $1$ ):

\frac{d}{d x} ((1 + x)^{1/2}) = \frac{1}{2} (1 + x)^{- 1/2} \cdot 1 = \frac{1}{2 1 + x}

Solved Example 5

Find the derivative of

x^{e}

(where

e

is Euler's number, a constant).

Solution:

Since $e$ is a real constant, the power rule applies directly:

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

Do not confuse $x^{e}$ with $e^{x}$ . Here $e$ is the exponent, not the base.

3. Fundamental Rules for Differentiation

The rules below let us differentiate any combination of standard functions. Throughout, $u$ and $v$ are differentiable functions of $x$ , and $c$ , $k$ are real constants.

(i) Derivative of a Constant

\frac{d}{d x} (c) = 0

Rate of change of a constant is zero.

(ii) Derivative of a Sum

If $y = u + v$ , then

\frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x}

The derivative of a sum is the sum of derivatives. This extends to any finite number of terms.

(iii) Derivative of a Difference

If $y = u - v$ , then

\frac{d y}{d x} = \frac{d u}{d x} - \frac{d v}{d x}

(iv) Constant Multiple Rule

\frac{d}{d x} (k u) = k \frac{d u}{d x}

(v) Product Rule

If $y = u \cdot v$ , then

\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}

For three factors $y = u v w$ :

\frac{d y}{d x} = u v \frac{d w}{d x} + u w \frac{d v}{d x} + v w \frac{d u}{d x}

(vi) Quotient Rule

If $y = \frac{u}{v}$ with $v \neq = 0$ , then

\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}

Solved Example 6

Find

\frac{d y}{d x}

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

Solution:

Split the expression first: $y = \frac{x ^{2}}{x} + \frac{1}{x} = x + \frac{1}{x} = x + x^{- 1}$ .

Using the sum rule and power rule:

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

Solved Example 7

Find

\frac{d y}{d x}

y = \frac{x ^{3} - 1}{x}

Solution:

Rewrite: $y = x^{2} - \frac{1}{x} = x^{2} - x^{- 1}$ .

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

Solved Example 8

Find

\frac{d y}{d x}

for each of the following:
(a)

y = (x + 2) (x + 3)

(b)

y = \frac{3 x + 4}{4 x + 5}

(c)

y = 1 + x^{2}

(d)

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

Solution:

(a) By the product rule with $u = x + 2$ , $v = x + 3$ :

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

(b) By the quotient rule with $u = 3 x + 4$ , $v = 4 x + 5$ :

\frac{d y}{d x} = \frac{( 4 x + 5 ) ( 3 ) - ( 3 x + 4 ) ( 4 )}{( 4 x + 5 ) ^{2}} = \frac{12 x + 15 - 12 x - 16}{( 4 x + 5 ) ^{2}} = \frac{- 1}{( 4 x + 5 ) ^{2}}

(c) Write $y = (1 + x^{2})^{1/2}$ and apply the power rule (with an implicit chain step for the inner $x^{2}$ ):

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

(d) Similarly $y = (a x^{2} + 2 b x + c)^{1/2}$ :

\frac{d y}{d x} = \frac{1}{2} (a x^{2} + 2 b x + c)^{- 1/2} (2 a x + 2 b) = \frac{a x + b}{a x ^{2} + 2 b x + c}

Common Mistakes to Avoid

Watch out

Product rule missed: writing $\frac{d}{d x} (uv) = \frac{d u}{d x} \cdot \frac{d v}{d x}$ is wrong. The derivative of a product is not the product of derivatives.
Quotient rule sign error: the numerator is $v \frac{d u}{d x} - u \frac{d v}{d x}$ , not $u \frac{d v}{d x} - v \frac{d u}{d x}$ . Order matters.
Confusing $x^{e}$ and $e^{x}$ : $\frac{d}{d x} (x^{e}) = e x^{e - 1}$ (power rule, constant exponent), whereas $\frac{d}{d x} (e^{x}) = e^{x}$ (constant base). Check which is variable before choosing the rule.
Forgetting to simplify first: for $y = \frac{x ^{2} + 1}{x}$ , splitting into $x + \frac{1}{x}$ is far faster than using the quotient rule.
Power rule and negative or fractional exponents: $\frac{d}{d x} (x) = \frac{d}{d x} (x^{1/2}) = \frac{1}{2} x^{- 1/2} = \frac{1}{2 x}$ . Do not treat $x$ as a special case; use the exponent form.
Constant vanishes, not the variable: $\frac{d}{d x} (5) = 0$ but $\frac{d}{d x} (5 x) = 5$ . The constant multiple rule keeps the constant.

Frequently Asked Questions

Q1. What does the derivative of a function represent geometrically?

The derivative $f^{'} (x_{0})$ equals the slope of the tangent line to the curve $y = f (x)$ at the point $(x_{0}, f (x_{0}))$ . Physically, if $x$ is time and $y$ is position, then $f^{'} (x)$ is instantaneous velocity. Whenever a quantity depends on another, the derivative gives the instantaneous rate of change of the first with respect to the second.

Q2. When does a derivative not exist?

A derivative fails to exist at a point if the limit $lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ does not exist finitely. Common cases: sharp corners (like $∣ x ∣$ at $x = 0$ ), vertical tangents (like $x^{1/3}$ at $0$ ), or points where the function is discontinuous. Differentiability implies continuity, but not the other way around.

Q3. Is the derivative from first principle needed for JEE Main and Advanced?

Yes. JEE Main and Advanced regularly ask you to differentiate a specific function from first principle, and understanding the limit definition is essential for continuity-differentiability questions.

Q4. Why is the power rule $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ true for all real $n$ ?

For integer $n$ it follows from the binomial theorem applied to $(x + h)^{n}$ . For rational and real $n$ , a rigorous proof uses logarithmic differentiation: write $y = x^{n}$ , take log to get $ln y = n ln x$ , differentiate to get $\frac{1}{y} y^{'} = \frac{n}{x}$ , so $y^{'} = \frac{n y}{x} = n x^{n - 1}$ .

Q5. What is the difference between $f^{'} (x)$ and $\frac{d y}{d x}$ ?

They mean the same thing when $y = f (x)$ . The notation $f^{'} (x)$ (Lagrange) emphasises the derivative as a new function; $\frac{d y}{d x}$ (Leibniz) emphasises the ratio of infinitesimal changes and makes chain rule and implicit differentiation more intuitive. Use whichever fits the problem better.

Q6. When should I use the product rule versus expanding first?

If the product is easy to expand (like $(x + 2) (x + 3)$ ), expand and then differentiate term by term - it is often faster. Use the product rule when expansion is messy or impossible, such as $x^{2} sin x$ or $e^{x} ln x$ . Both routes give the same answer; pick the shorter one.

Q7. Can the derivative be negative or zero?

Absolutely. A negative derivative means the function is decreasing at that point (tangent slopes downward). A zero derivative means the tangent is horizontal - this happens at maxima, minima and points of inflection. These sign properties are the basis of the increasing-decreasing test in JEE application-of-derivative problems.

Q8. How many standard derivatives should I memorise for JEE?

Twelve is the minimum: $x^{n}$ , constant, $e^{x}$ , $a^{x}$ , $ln x$ , $lo g_{a} x$ , and the six trigonometric functions ( $sin$ , $cos$ , $tan$ , $cot$ , $sec$ , $csc$ ). Add the six inverse trigonometric derivatives (covered in the next concept) and you can differentiate any exam-level expression when combined with the chain, product and quotient rules.

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1. Derivative from the First Principle

Geometric Meaning: Slope of the Tangent

2. Standard Derivatives of Important Functions

3. Fundamental Rules for Differentiation

(i) Derivative of a Constant

(ii) Derivative of a Sum

(iii) Derivative of a Difference

(iv) Constant Multiple Rule

(v) Product Rule

(vi) Quotient Rule

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What does the derivative of a function represent geometrically?

Q2. When does a derivative not exist?

Q3. Is the derivative from first principle needed for JEE Main and Advanced?

Q4. Why is the power rule dxd​(xn)=nxn−1 true for all real n?

Q5. What is the difference between f′(x) and dxdy​?

Q6. When should I use the product rule versus expanding first?

Q7. Can the derivative be negative or zero?

Q8. How many standard derivatives should I memorise for JEE?

Ready to master Limits And Derivatives?

Q4. Why is the power rule $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ true for all real $n$ ?

Q5. What is the difference between $f^{'} (x)$ and $\frac{d y}{d x}$ ?