What is the derivative of a constant value?

The derivative of constant function is zero.

What are the application of differentiation?

We use the differentiation to find the maximum or minimum values of a function, the velocity and acceleration of moving objects and the tangent of the curve.

Home

JEE Maths

Differentiation

Frequently Asked Questions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Differentiation

Differentiation is a mathematical process that measures how a function changes as its input changes. In simpler terms, it calculates the rate of change of function.

The derivative can be visualized as the slope of the tangent line to the functions graphs at the point.

Differentiation is used to calculate rates of change in various field, such as physics (velocity, acceleration…..)

1.0Derivative By First Principle

Let y = f(x) :

y + Δy = f(x + Δx)

$∴ \frac{Δ y}{Δ x} = \frac{f ( x + Δ x ) - f ( x )}{Δ x}$ (average rate of change of function)

$∴ \frac{dy}{dx} = lim_{Δ x \to 0} \frac{Δ y}{Δ x} = lim_{Δ x \to 0} \frac{f ( x + Δ x ) - f ( x )}{Δ x}$ …..(i)

(i) denotes the instantaneous rate of change of function and gives slope of tangent at any point on curve y = f(x). Finding the value of the limit given by (i) in respect to a variety of functions is called finding the derivative by first principle/by delta method/by ab-initio method/by fundamental definition of calculus.

Note than if y = f(x) then the symbols $\frac{d y}{d x} = Dy = f^{'} (x) = y_{1}$ or y’ have the same meaning.

Derivative of standard functions:

f(x)	f’(x)
xⁿ	nx^n-1
e^x	e^x
a^x	a^x lna, a > 0
lnx	1/x
log_ax	(1/x) log_ae, a > 0, a ≠ 1
sinx	cosx
cosx	– sinx
tanx	sec²x
secx	secx tanx
cosecx	–cosec²x
constant	0
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	$\frac{1}{1 + x ^{2}}, x \in R$
sec^-1 x	$\frac{1}{∣ x ∣ x ^{2} - 1}, ∣ x ∣ > 1$
cosec^-1 x	$\frac{- 1}{∣ x ∣ x ^{2} - 1}, ∣ x ∣ > 1$
cot^-1 x	$\frac{- 1}{1 + x ^{2}}, x \in R$

2.0Fundamental Theorems

If f and g are derivable functions of x, then,

(a) $\frac{d}{dx} (f \pm g) = \frac{df}{dx} \pm \frac{dg}{dx}$

(b) $\frac{d}{d x} (c f) = c \frac{df}{d x}$ , where c is any constant

(c) $\frac{d}{d x} (f g) = f \frac{d g}{d x} + g \frac{df}{d x}$ known as “PRODUCT RULE”

(d) $\frac{d}{d x} (\frac{f}{g}) = \frac{g ( \frac{df}{dx} ) - f ( \frac{dg}{dx} )}{g ^{2}}$ where g ≠ 0 known as “QUOTIENT RULE”

(e) If y = f(u) & u = g (x) then $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ known as “CHAIN RULE”

Note : in general if y = f(u) then $\frac{dy}{dx} = f^{'} (u) \cdot \frac{du}{dx}$

Logarithmic differentiation:

To find the derivative of a function:

(a) which is the product of quotient of a number of functions or

(b) of the form [f(x)]^g(x) where f and g are both derivable functions.

It is convenient to take the logarithm of the function first and then differentiate.

Parametric differentiation:

If y = f(θ) and x = g(θ) where θ is a parameter, then $\frac{d y}{d x} = \frac{d y / d θ}{d x / d θ} = \frac{f ^{'} ( θ )}{g ^{'} ( θ )}$

Derivative of a function w.r.t. another function

Let y = f(x) ; z = g(x) then $\frac{d y}{d z} = \frac{d y / d x}{d z / d x} = \frac{f ^{'} ( x )}{g ^{'} ( x )}$

Differentiation of implicit functions: φ(x, y) = 0

To find $\frac{d y}{d x}$ of implicit functions, we differentiate each term w.r.t. regarding y as a function of x and then collect terms with $\frac{dy}{dx}$ together on one side.
Also $\frac{dy}{dx} = \frac{- \frac{\partial ϕ}{\partial x}}{\frac{\partial ϕ}{\partial y}}$ , where $\frac{\partial ϕ}{\partial y}$ = partial derivative of φ(x, y) w.r.t. x taking y as a constant and $\frac{\partial ϕ}{\partial y}$ = partial derivative of φ(x, y) w.r.t. y taking x as a constant. (In the case of implicit functions, generally, both x and y present in answers of dy/dx)

3.0Derivative Of Inverse Function

Theorem: If the inverse functions f and g are defined by y = f(x) and x = g(y) and if f'(x) exists and f’(x) ≠ 0, then $g ’ (y) = \frac{1}{f ^{'} ( x )}$ . This result can also be written as, if $\frac{dy}{dx}$ exist and $\frac{d y}{d x} \neq = 0$ , then $\frac{d x}{d y} = \frac{1}{\frac{d y}{d x}}$ or $\frac{d y}{d x} \cdot \frac{d x}{d y} = 1$ or $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} [\frac{dx}{dy} \neq = 0]$

Note: (i) $g ’ (f (x)) = \frac{1}{f ^{'} ( x )}$ (ii) $g^{''} (f (x)) = \frac{- f ^{''} ( x )}{( f ^{'} ( x ) ) ^{3}}$

Higher order Derivatives

Let a function y = f(x) be defined on an interval (a, b). If f(x) is differentiable function, then its derivative f’(x) [or (dy/dx) or y’] is called the first derivative of y w.r.t. x if f’(x) is again differentiable function on (a, b), then its derivative f”(x) [or d²y/dx² or y”] is called second derivative of y w.r.t. x similarly, the 3^rd order derivative of y w.r.t. x, if exists, is defined by $\frac{d ^{3} y}{dx ^{3}} = \frac{d}{dx} (\frac{d ^{2} y}{dx ^{2}})$ and denoted by f”’ (x) or y”’ and so on.

Note: If x = f (θ) and y = g(θ) where ‘θ’ is a parameter then $\frac{d y}{d x} = \frac{d y / d θ}{d x / d θ} & \frac{d ^{2} y}{d x ^{2}} = \frac{d}{d θ} (\frac{d y}{d x}) / \frac{d x}{d θ}$ , In general $\frac{d ^{n} y}{d x ^{n}} = \frac{d}{d θ} (\frac{d ^{n - 1} y}{d x ^{n - 1}}) / \frac{d x}{d θ}$

L. Hopital’s Rule

(a) This rule is applicable for the indeterminate forms of the type $\frac{0}{0}, \frac{\infty}{\infty}$ . If the function f(x) and g(x) are differentiable in certain neighbourhood of the point ‘a’, except, may be, at the point ‘a’ itself and g'(x) ≠ 0, and if

$lim_{x \to a} f (x) = lim_{x \to a} g (x) = 0$

or $lim_{x \to a} f (x) = lim_{x \to a} g (x) = \infty$ ,

then $lim_{x \to a} \frac{f ( x )}{g ( x )} = lim_{x \to a} \frac{f ^{'} ( x )}{g ^{'} ( x )}$

provided the limit $lim_{x \to a} \frac{f ^{'} ( x )}{g ^{'} ( x )}$ exists (L’ Hopital’s rule). The point ‘a’ may be either finite or improper ( $+ \infty or - \infty$ )

(b) Indeterminate forms of the type 0. ∞ or ∞ – ∞ are reduced to forms of the type $\frac{0}{0}$ or $\frac{\infty}{\infty}$ by algebraic transformations.

(c) Indeterminate forms of the type 1^∞, ∞⁰ or 0⁰ are reduced to forms of the type 0 × ∞ by taking logarithms or by the transformation $[f (x)]^{ϕ (x)} = e^{ϕ (x) \cdot l n f (x)}$

4.0Solved Examples on Differentiation

Ex.1 Differentiate each of following functions by first principle:

(i) f(x) = tanx (ii) f(x) = e^sinx

Sol.

(i) $f^{'} (x) = lim_{h \to 0} \frac{t a n ( x + h ) - t a n x}{h} = lim_{h \to 0} \frac{t a n ( x + h - x ) [ 1 + t a n x t a n ( x + h )]}{h}$

$= f^{'} (x) = lim_{h \to 0} \frac{t a n h}{h} \cdot (1 + tan^{2} x) = sec^{2} x$ .

$= f^{'} (x) = lim_{h \to 0} \frac{t a n h}{h} \cdot (1 + tan^{2} x) = sec^{2} x .$

(ii) $f^{'} (x) = lim_{h \to 0} \frac{e ^{s i n (x + h)} - e ^{s i n x}}{h} = lim_{h \to 0} e^{s i n x} \frac{[ e ^{s i n (x + h) - s i n x} - 1 ]}{s i n ( x + h ) - s i n x} (\frac{s i n ( x + h ) - s i n x}{h})$

$= e^{s i n x} lim_{h \to 0} \frac{s i n ( x + h ) - s i n x}{h} = e^{s i n x} cos x$

Ex.2 Differentiate each of following functions

(i) y = x² + sin x (ii) y = e^x tan x

(iii) $y = \frac{x}{s i n x}$ (iv) y = cos (lnx)

Sol.

(i) y = x² + sin x

$\frac{d y}{d x} = \frac{d}{d x} (x^{2} + sin x) = \frac{d ( x ^{2} )}{d x} + \frac{d ( s i n x )}{d x}$

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

(ii) y = e^x tan x

$\frac{d y}{d x} = \frac{d}{d x} (e^{x} tan x) = tan x \cdot \frac{d ( e ^{x} )}{d x} + e^{x} \frac{d}{d x} (tan x)$

$\frac{d y}{d x} = t an x . e x + e x sec 2 x$

(iii) $y = \frac{x}{s i n x}$

$\frac{d y}{d x} = \frac{s i n x \frac{d}{d x} ( x ) - x \frac{d}{d x} ( s i n x )}{s i n ^{2} x}$

$\frac{d y}{d x} = \frac{s i n x - x c o s x}{s i n ^{2} x}$

(iv) y = cos (lnx)

$\frac{dy}{dx} = \frac{d}{dx} (cos (l n x))$

$= - s in (l n x) . \frac{d}{dx} (l n x)$

$= - s in (l n x) \times \frac{1}{x}$

$= - \frac{s i n ( l n x )}{x}$

Ex.3 If $y = lo g_{e} (tan^{- 1} 1 + x^{2})$ , find $\frac{dy}{dx} .$

Sol.

$y = lo g_{e} (tan^{- 1} 1 + x^{2})$

On differentiating we get.

$= \frac{1}{t a n ^{- 1} 1 + x ^{2}} \cdot \frac{1}{1 + ( 1 + x ^{2} ) ^{2}} \cdot \frac{1}{2 1 + x ^{2}} \cdot 2 x$

$= \frac{x}{( t a n ^{- 1} 1 + x ^{2} ) { 1 + ( 1 + x ^{2} ) ^{2} } 1 + x ^{2}} = \frac{x}{( t a n ^{- 1} 1 + x ^{2} ) ( 2 + x ^{2} ) 1 + x ^{2}}$

Ex.4 Find $\frac{dy}{dx}$ if y = x^x

Sol.

y = x^x

Taking log on both side, we get

log y = log (x^x)

log y = x log x

on differentiating, we get

$\frac{1}{y} \frac{d y}{d x} = lo g x + x \times \frac{1}{x}$

$\frac{d y}{d x} = y [lo g x + 1]$

$\frac{d y}{d x} = x^{x} [lo g x + 1]$

Ex.5 If $y = \frac{x ^{1/2} ( 1 - 2 x ) ^{2/3}}{( 2 - 3 x ) ^{3/4} ( 3 - 4 x ) ^{4/5}}$ find $\frac{dy}{dx}$

Sol.

ln $y = \frac{1}{2} l n x + \frac{2}{3} l n (1-2 x) - \frac{3}{4} l n (2-3 x) - \frac{4}{5} l n (3-4 x)$

On differentiating we get,

$\Rightarrow \frac{1}{y} \frac{d y}{d x} = \frac{1}{2 x} - \frac{4}{3 ( 1 - 2 x )} + \frac{9}{4 ( 2 - 3 x )} + \frac{16}{5 ( 3 - 4 x )}$

$\frac{d y}{d x} = y (\frac{1}{2 x} - \frac{4}{3 ( 1 - 2 x )} + \frac{9}{4 ( 2 - 3 x )} + \frac{16}{5 ( 3 - 4 x )})$

Ex.6 If x^y + y^x = 2, then find $\frac{dy}{dx}$ .

Sol.

Let u = x^y and v = y^x

u + v = 2 ⇒ $\frac{d u}{d x} + \frac{d v}{d x} = 0$

Now u = x^y and v = y^x

⇒ ln u = y lnx and ln v = x ln y

$\frac{1}{u} \frac{d u}{d x} = \frac{y}{x} + ln x \frac{d y}{d x}$ and

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

$\Rightarrow \frac{d u}{d x} = x^{y} (\frac{y}{x} + ln x \frac{d y}{d x}) an d \frac{d v}{d x} = y^{x} (ln y + \frac{x}{y} \frac{d y}{d x})$

$\Rightarrow x^{y} (\frac{y}{x} + ln x \frac{d y}{d x}) + y^{x} (ln y + \frac{x}{y} \frac{d y}{d x}) = 0$

$\Rightarrow \frac{d y}{d x} = - \frac{( y ^{x} l n y + x ^{y} \cdot \frac{y}{x} )}{( x ^{y} l n x + y ^{x} \cdot \frac{x}{y} )}$

Aliter :

φ(x, y) = x^y + y^x – 2 = 0

$\frac{dy}{dx} = \frac{\partial ϕ / \partial x}{\partial ϕ / \partial y} = - (\frac{yx ^{n - 1} + y ^{x} l n y}{x ^{y} l n x + xy ^{x - 1}})$

Ex.7 Differentiation log_e (tan x) with respect to sin^–1(e^x)

Sol.

$\frac{d ( l o g _{e} t a n x )}{d ( s i n ^{- 1} ( e ^{x} )} = \frac{c o t x \cdot s e c ^{2} x}{e ^{x} \cdot 1/ 1 - e ^{2 x}} = \frac{e ^{- x} 1 - e ^{2 x}}{s i n x c o s x}$

Ex.8 If f(x) = 2^x + x³ + 1 and g(x) = f^–1(x), then find g’(2)

Sol.

$g (x) = f -1 (x)$

$g (f (x)) = x$

$g ’ (f (x)) . f ’ (x) = 1$

$g ’ (f (x) = \frac{1}{f ^{'} ( x )} = \frac{1}{2 ^{x} l n 2 + 3 x ^{2}}$

$f (x) = 2$

$2^{x} + x^{3} + 1 = 2 \Rightarrow x = 0$

$g^{'} (f (0)) = \frac{1}{f ^{'} ( 0 )} = \frac{1}{l n 2}$

$g^{'} (2) = \frac{1}{l n 2}$

Ex.9 If x = a(t + sin t) and y = a(1 – cos t), find $\frac{d ^{2} y}{d x ^{2}}$ .

Sol.

Here x = a(t + sin t) and y = a(1 – cos t)

Differentiating both sides w.r.t., t we get :

$\frac{dx}{dt} = a (1 + cos t)$ and $\frac{d y}{d t} = a (sin t)$

$∴ \frac{dy}{dt} = \frac{a s i n t}{a ( 1 + c o s t )} = \frac{2 s i n \frac{t}{2} \cdot c o s \frac{t}{2}}{2 c o s ^{2} \frac{t}{2}} = tan (\frac{t}{2})$

Again differentiating both sides, we get,

$\frac{d ^{2} y}{dx ^{2}} = sec^{2} (\frac{t}{2}) \cdot \frac{1}{2} \cdot \frac{dt}{dx} = \frac{1}{2} sec^{2} (t /2) \cdot \frac{1}{a ( 1 + c o s t )} = \frac{1}{2 a} \cdot \frac{s e c ^{2} ( \frac{t}{2} )}{2 ( c o s ^{2} \frac{t}{2} )}$

Hence, $\frac{d ^{2} y}{dx ^{2}} = \frac{1}{4 a} \cdot sec^{4} (\frac{t}{2})$

Ex.10 Find $lim_{x \to 0} \frac{s i n x - x}{x ^{3}}$

Sol.

$lim_{x \to 0} \frac{s i n x - x}{x ^{3}} (\frac{0}{0})$

applying L’ Hopital rule

$lim_{x \to 0} \frac{c o s x - 1}{3 x ^{2}} = lim_{x \to 0} - \frac{1}{3} \frac{( 1 - c o s x )}{x ^{2}} = - \frac{1}{6}$

5.0Practice Problem on Differentiation

(1) Find the derivative of y = cos(x²) using first principle.

(2) If f(x) = (1 + x)(3 + x²)^1/2, then f'(–1) is equal to-

(A) 0 (B) $22$ (C) 4 (D) 6

(3) If $y = \frac{x ^{12} + x ^{6} + 1}{x ^{6} + x ^{3} + 1} & \frac{d y}{d x} = x^{2} (a x^{3} + b)$ , then the value of (a + b) is equal to

(4) If f(x) = (x + 1)(x + 2) ……. (x + n) the f'(0) is

(5) If x = secθ & y = secⁿθ – cosⁿθ, then show that $(x^{2} + 4) (\frac{d y}{d x})^{2} = n^{2} (y^{2} + 4)$

(6) If $y = sin x + sin x + sin x + \dots\dots \infty$ , then $\frac{d y}{d x} (sin x > 0)$ , is equal to-

(A) $\frac{c o s x}{2 y - 1}$ (B) $\frac{c o s x}{1 + 4 s i n x}$

(C) $\frac{y c o s x}{y ^{2} + s i n x}$ (D) $\frac{y c o s x}{y ^{2} - s i n x}$

(7) If x = 2cos t – cos 2t ; y = 2sin t – sin 2t, find $\frac{d ^{2} y}{dx ^{2}}$ at $t = \frac{π}{2} .$

(8) $\frac{d}{dx} {sin^{2} (cot^{- 1} \frac{1 + x}{1 - x})} =$

(A) $- \frac{1}{2}$ (B) 0 (C) $\frac{1}{2}$ (D) –1

(9) If f(x) = x + 3x³ + 5x⁵ and g = f^–1, then find g'(9) and g"(9)

(10) Prove that $lim_{x \to 0^{+}} x^{x} = 1$

Answers:

(1) –2x sin(x²)

(2) C

(3) 3

(4) $n! (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \dots .. + \frac{1}{n})$

(6) A, B, C

(7) $- \frac{3}{2}$

(8) A

(9) $g^{'} (9) = \frac{1}{35}, g^{''} (9) = \frac{118}{( 35 ) ^{3}}$

6.0Sample Questions on Differentiation

Q.1 What is differentiation?

Ans. The instantaneous rate of change of a function with respect to another quantity is called differentiation.

If y = f(x) is a differentiable function of x, then

$\frac{d y}{d x} = f^{'} (x) = lim_{Δ x \to 0} \frac{f ( x + Δ x ) - f ( x )}{Δ x}$

Q.2 What are the differentiation Rules in calculus?

Ans. If f and g are derivable functions of x, then

(a) $\frac{d}{dx} (f \pm g) = \frac{df}{dx} \pm \frac{dg}{dx}$

(b) $\frac{d}{dx} (cf) = c \frac{df}{dx}$ , where c is any constant

(c) $\frac{d}{d x} (f g) = f \frac{d g}{d x} + g \frac{df}{d x}$ where g ≠ 0 known as “PRODUCT RULE”

(d) $\frac{d}{dx} (\frac{f}{g}) = \frac{g ( \frac{df}{dx} ) - f ( \frac{dg}{dx} )}{g ^{2}}$ where g ≠ 0 known as “QUOTIENT RULE”

(e) If y = f(u) & u = g(x) then $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ known as “CHAIN RULE”

Frequently Asked Questions

What is the derivative of a constant value?

What are the application of differentiation?

Frequently Asked Questions

What is the derivative of a constant value?

What are the application of differentiation?

Join ALLEN!

Join ALLEN!

Differentiation

1.0Derivative By First Principle

2.0Fundamental Theorems

3.0Derivative Of Inverse Function

Higher order Derivatives

L. Hopital’s Rule

4.0Solved Examples on Differentiation

5.0Practice Problem on Differentiation

6.0Sample Questions on Differentiation

Table of Contents