# Differential Coefficient

Differential coefficient of a function y = f(x) at x = a is f’(a). As far as the JEE exam is concerned, differential coefficient is an important topic. Students can expect questions from this topic for any competitive exams. In this article, we discuss the concept of differential coefficient and the conditions under which it exists.

Consider the function y = f(x). The differential coefficient of y at x = a is f’(a). f’(a) will exist only when LHD = RHD at x = a. When LHD = RHD at x = a, then f’(a) = f’(a) = f’(a+).

Let us consider a = 5. If we have to find out f’(5), then we can say that f’(5) exists only if f’(5) = f’(5+).

f’(5) = f’(5) = f’(5+)

f’(5) = limh→ 0[f(5+h) – f(5)]/h = limh→ 0[f(5-h) – f(5)]/-h

In general

• f’(a+) = limh→ 0[f(a+h) – f(a)]/h = f’(a)
• f’(a) = limh→ 0[f(a-h) – f(a)]/-h = f’(a)
• f’(a) = limx→ a[f(x) – f(a)]/(x-a)

### Differential Coefficient of Standard Functions

(1) d/dx (k) = 0, k is a constant

(2) d/dx (xn) = nxn-1

(3) d/dx (log x) = 1/x

(4) d/dx (ex) = ex

(5) d/dx (ax) = ax loge a

(6) d/dx (sin x) = cos x

(7) d/dx (cos x) = -sin x

(8) d/dx (tan x) = sec2 x

### Solved Examples

Example 1: The differential coefficient of the given function $\log_{e}\left ( \sqrt{\frac{1+\sin x}{1-\sin x}} \right )$ is

(1) cosec x

(2) tan x

(3) sec x

(4) cos x

Solution:

Let y = loge (√((1 + sin x)/(1 – sin x))

Multiply numerator and denominator by √(1 + sin x)

=> loge (√((1 + sin x)2/(1 – sin2 x))

= loge (√((1 + sin x)2/cos2 x)

= loge ((1 + sin x)/cos x)

= loge(sec x + tan x)

dy/dx = [1/(sec x + tan x)] × sec x tan x + sec2x

= [1/(sec x + tan x)] × sec x (tan x + sec x)

= sec x

Hence option (3) is the answer.

Example 2: The differential coefficient of x6 with respect to x3 is

(1) 8x2

(2) 3x3

(3) 5x3

(4) 2x3

Solution:

Let u = x6

Differentiate with respect to x we get

du/dx = 6x5

Let v = x3

dv/dx = 3x2

du/dv = (du/dx)/(dv/dx)

= 6x5/3x2

= 2x3

Hence option (4) is the answer.

Example 3: Find the differential coefficient of the function x2 log x sin x with respect to x.

(1) x2 log x cos x + x sin x + 2x log x sin x

(2) x log x cos x + x sin x – 2x log x sin x

(3) x2 log x cos x + x sin x + 2x2 log x sin x

(4) x2 log x cos x + x2 sin x + 2x log x sin x

Solution:

Let y = x2 log x sin x

dy/dx = x2 d/dx (log x sin x) + 2x log x sin x

= x2 log x cos x + x2 sin x (1/x) + 2x log x sin x

= x2 log x cos x + x sin x + 2x log x sin x

Hence option (1) is the answer.

Example 4: If y = sin [ cos (sin x)], then dy/dx =

(1) -cos [cos (sin x)] sin (cos x) cos x

(2) -cos [cos (sin x)] sin (sin x) cos x

(3) cos [cos (sin x)] sin (cos x) cos x

(4) cos [cos (sin x)] sin (sin x) cos x

Solution:

Given y = sin [ cos (sin x)]

dy/dx = cos [ cos (sin x)] ×(d/dx) (cos sin x)

= cos [ cos (sin x)] ×(-sin (sin x) cos x

= – cos [cos (sin x)] sin (sin x) cos x

Hence option (2) is the answer.

Example 5: Find the differential coefficient of sin 2nx/cos2 nx

(1) 2n cosec2 nx

(2) 2n sec2 nx

(3) 2n2 sec2 nx

(4) 2n2 tan2 nx

Solution:

Let y = sin 2nx/cos2 nx

= 2 sin nx cos nx/cos2 nx

= 2 sin nx/cos nx

= 2 tan nx

Differentiate with respect to x

dy/dx = 2 sec2 nx . n

= 2n sec2 nx

Hence option (2) is the answer.

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