0412/IIT.14/CR/Bk.4/Ch.19/Pg.80

Derivatives

1) Concept of Differentiability (or first principle of differentiation)

A function f(x), x R, is said to be differentiable at same point, say x = a if the right hand derivative and the left hand derivative,

i.e., h 0 h 0

f (a h) f (a) f (a h) f (a)lim and lim

h h

, respectively, both exists and are equal. If

L.H.D. = R.H.D., then we say that the function is differentiable at some point, say x = a and the

derivative at that point (x = a) is given by,

f (a) = h 0 h 0

f (a h) f (a) f (a h) f (a)lim lim

h h

and common value denoted by f (a) or d

[f (x)]dx

at x = a.

2) Geometrical meaning of the derivative

Let us consider a function y = f (x) and a point P (x, f (x)) on the curve. Consider a point

A (x + h, f (x + h)) on the right hand side of the point (x, f (x)).

Now, as h 0

PA 0

(i.e., the distance PA tends to zero or to a single point P)

h 0Lim

(Slope of chord PA) (Slope of tangent at P)

h 0

f (x h) f (x)Lim

h

= slope of tangent at P

Which means that the value of the derivative f '(x) for a given value of x is equal to the slope of

the tangent to the curve y = f (x) at the point P (x, y).

Note that :

1. A function f(x) is said to be nondifferentiable at a point x = a, if (a) One or both, the left handed and right handed derivatives, do not exist at x = a.

(b) Both the derivatives L.H.D. and R.H.D. exist but are unequal.

2. Every function which has a finite derivative at x = a must be continuous at x = a. However, the

converse is not true i.e., if a function f(x) is continuous at x = a, then it may or may not be

differentiable at this point. For example, |x| is continuous everywhere but it is not differentiable at

x = 0 because left hand derivative and right hand derivative are not equal.

Thus, continuity is necessary but not a sufficient condition for differentiability.

Answer keys for objectives to be changed in every chapter.

x h

x

y = f(x)

(x, f(x))

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

P

A

Notes on Derivatives (81)

0412/IIT.14/CR/Bk.4/Ch.19/Pg.81

3) Derivatives in Closed Interval

Let f(x) be defined on [a, b], then it is said to be differentiable on [a, b] if it is differentiable at

each point of (a, b) and from the right at a and left at b (i.e., x a

f (x) f (a)lim

x a

and

x b

f (x) f (b)lim

x b

) both exists.

4) Some standard formulae and properties

If f(x) and g(x) are differentiable functions, then

i) d d d

f (x) g(x) f (x) g(x)dx dx dx

ii) d d

k(f (x) k f (x)dx dx

, where k is a constant.

iii) Product rule :

d d d

f (x).g(x) g(x) f (x) f (x) g(x)dx dx dx

iv) Quotient rule :

2

d dg(x) f (x) f (x) g(x)

d f (x) dx dx

dx g(x) g(x)

v) Chain rule :

If y = f(u) and u = g(x), then

dy dy du

.dx du dx

Or, in other words, if, y = f (g(x)) (, u = g(x))

then dy

f '(g(x)).g '(x)dx

.

5) Differentiation of Some Elementary Functions

(i) The trigonometric functions have the following derivatives :

d

(sin x)dx

= cos x d

(cos x)dx

= sin x

d

(tan x)dx

= 2sec x d

(cot x)dx

= 2cosec x

d

(sec x)dx

= sec x tan x d

(cosec x)dx

= cosec x cot x

(ii) If f (x) is a differentiable function

n n 1d

(x ) nxdx

n R, x > 0 and in general

n n 1d f (x) n (f (x)) f (x)dx

.

(iii) If f (x) is a differentiable function

d 1

(ln x)dx x

and in general d 1

(ln f (x)) f (x)dx f (x)

(82) Vidyalankar : IIT Maths

0412/IIT.14/CR/Bk.4/Ch.19/Pg.82

(iv) x xd

(a ) a ln adx

. In particular, x xd

(e ) edx

.

(v) 1 1

2

d d 1(sin x) (cos x)

dx dx 1 x

for 1 < x < 1.

At the points x = 1, 1 1sin x and cos x are not differentiable.

(vi) 1 12

d d 1(tan x) (cot x)

dx dx 1 x

for x R

(vii) 1 1

2

d d 1(sec x) (cosec x)

dx dx | x | x 1

for | x | > 1.

6) Derivative of Parametric Equations

If x = f (t) and y = g (t), then g (t)dy dy dx

dt dtdx f (t)

.

7) Derivative of Implicit Functions

If f (x, y) = 0, then on differentiating of f (x, y) w.r.t. x, we get d

f (x, y)dx

= 0. Collect the

terms of dy

dx and solve.

Alternate Method :

dy f f

x ydx

In particular,

if 1 2 3 nf (x ,x ,x , ......, x ) 0 and 2 3 nx ,x , ....., x are the functions of 1x then

1

df

dx = 32 n

1 2 1 3 1 n 1

dxdx dxf f f f. . ..... .

x x dx x dx x dx

8) Some Standard Substitutions

Expression Substitution

2 2a x x = a sin or a cos

2 2a x x = a tan or a cot

2 2x a x = a sec or a cosec

a x

a x

or

a x

a x

x = a cos or a cos 2

2(2ax x ) x = a (1 cos )

Notes on Derivatives (83)

0412/IIT.14/CR/Bk.4/Ch.19/Pg.83

9) Partial Derivatives

Partial derivative of f (x1, x2, , xn) with respect to x1 means derivative of f (x1, x2, , xn) with respect to x1 treating all other variables x2, , xn as constants. It is

represented by 1

' 'f

x

.

Also 1

df

dx = 1 2 n

1 1 2 1 n 1

dx dx dxf f f. . ... .

x dx x dx x dx

In particular, if f (x, y) = 0, then

d f (x, y)

dx =

f dx f dy. .

x dx y dx

= 0

dy

dx =

f f

x y

10) Logarithmic Differentiation :

It is used for differentiating functions of the type, y = (x){f (x)} .

Method 1 : y = (x){f (x)}

lny = (x) ln{f (x)}

1 dy

y dx =

(x)f '(x) '(x) ln{f (x)}

f (x)

dy

dx = (x)

(x) f '(x){f (x)} '(x) ln{f (x)}

f (x)

Method 2 : y = (x){f (x)}

y = (x) ln{f (x)}e ( x = ln xe )

dy

dx = (x)ln{f (x)}

(x)e f '(x) '(x) ln{f (x)}

f (x)

dy

dx = (x)

(x) f '(x){f (x)} '(x) ln{f (x)}

f (x)

Method 3 : (Using partial derivatives)

If y = (x){f (x)} , then

dy

dx = derivative of {f (x)}

(x) w.r.t. x taking (x) as a constant

+ derivative of {f(x) (x)

w.r.t. x taking f(x) as a constant

dy

dx = (x) 1 (x)(x) {f (x)} f '(x) {f (x)} ln{f (x)}. '(x)

dy

dx = (x)

(x) f '(x){f (x)} '(x) ln (f (x))

f (x)

x

y

P (x, f(x))

(84) Vidyalankar : IIT Maths

0412/IIT.14/CR/Bk.4/Ch.19/Pg.84

11) Higher Derivative of Function

1. If y = f (x) then the derivative of dy

dx w.r.t. x is called the second derivative of y w.r.t. x and it is

denoted by 2

2

d y

dx.

Also, 2

2

d y d dy

dx dxdx

;

3 2

3 2

d y d d y

dxdx dx

2. If y as a function of x is given in parametric form by y = (t) and x = (t), then

dy '(t)

dx '(t)

and 2

2

d y

dx =

d dy

dx dx

= d dy dt

dt dx dx

=

d '(t) 1

dt '(t) '(t)

2

2

d y

dx =

3

'(t) ''(t) '(t) ''(t)

( '(t))

Note : Don't write, 2

2

d y

dx =

2 2

2 2

d y dt

d x dt =

''(t)

''(t)

[

2 2

2 2

d y d y

dx d x ]