# Using differentials, find the approximate value of each of the following up to 3 places of decimal

(i) √25.3 (ii) √49.5 (iii) √0.6

(iv) (0.009)^{1/3} (v) (0.999)^{1/10} (vi) (15)^{1/4}

(vii) (26)^{1/3} (viii) (255)^{1/4} (ix) (82)^{1/4}

(x) (401)^{1/2} (xi) (0.0037)^{1/2} (xii) (26.57)^{1/3}

(xiii) (81.5)^{1/4} (xiv) (3.968)^{3/2} (xv) (32.15)^{1/5}

**Solution:**

We can use differentials to calculate small changes in the dependent variable of a function corresponding to small changes in the independent variable

(i) √25.3

Consider y = √x

Let x = 25 and Δx = 0.3

Then,

Δy = √x + Δx - √x

= √25.3 - √25

= √25.3 - 5

Δy + 5 = √25.3

Now, dy is approximately equal to Δy and is given by,

dy = (dy/dx) Δx

= 1/2√x (0.3) [∵ y = √x]

= 1/2√25 (0.3)

= 0.03

Hence,

√25.3 = 0.03 + 5

= 5.03

Thus, the approximate value of √25.3 = 5.03.

(ii) √49.5

Consider y = √x

Let x = 49 and Δx = 0.5

Then,

Δy = √x + Δx - √x

= √49.5 - √49

= √49.5 - 7

Δy + 7 = √49.5

Now, dy is approximately equal to Δy and is given by,

dy = (dy/dx) Δx

= 1/2√x (0.5) [∵ y = √x]

= 1/2√49 (0.5)

= 0.035

Hence,

√49.5 = 7 + 0.035

= 7.035

Thus, the approximate value of √49.5 = 7.035.

(iii) √0.6

Consider y = √x

Let x = 1 and Δx = - 0.4

Then,

Δy = √x + Δx - √x

= √0.6 - √1

Δy + 1 = √0.6

Now, dy is approximately equal to Δy and is given by,

dy = (dy/dx) Δx

= 1/2√x (- 0.4) [∵ y = √x]

= 1/2 (- 0.4)

= - 0.2

Hence,

√0.6 = 1 + (- 0.2)

= 1- 0.2

= 0.8

Thus, the approximate value of √0.6 = 0.8.

(iv) (0.009)^{1/3}

Consider y = (x)^{1/3}

Let x = 0.008 and Δx = 0.001

Then,

Δy = (x + Δx)^{1/3} - (x)^{1/3}

= (0.009)^{1/3} - (0.008)^{1/3}

Δy + 0.02 = (0.009)^{1/3}

Now, dy is approximately equal to Δy and is given by,

dy = (dy/dx) Δx

= 1/3(x)^{2/3} Δx [∵ y = (x)^{1/3}]

= [1/(3 x 0.04)] (0.001)

= 0.001/0.12

= 0.008

Hence,

(0.009)^{1/3 }= 0.2 + 0.008

= 0.208

Thus, the approximate value of (0.009)^{1/3 }= 0.208.

(v) (0.999)^{1/10}

Consider y = (x)^{1/10}

Let x = 1 and Δx = - 0.001

Then,

Δy = (x + Δx)^{1/10} - (x)^{1/10}

= (0.999)^{1/10} - 1

Δy + 1 = (0.999)^{1/10}

Now, dy is approximately equal to Δy and is given by,

dy = (dy/dx) Δx

= 1/10(x)^{9/10} Δx [∵ y = (x)^{1/10}]

= 1/10 (- 0.001)

= - 0.0001

Hence,

(0.999)^{1/10} = 1 + (- 0.0001)

= 0.9999

Thus, the approximate value of (0.999)^{1/10}^{ }= 0.9999.

(vi) (15)^{1/4}

Consider y = (x)^{1/4}

Let x = 16 and Δx = - 1

Then,

Δy = (x + Δx)^{1/4} - (x)^{1/4}

= (15)^{1/4} - (16)^{1/4}

= (15)^{1/4} - 2

Δy + 2 = (15)^{1/4}

Now, dy is approximately equal to Δy and is given by,

dy = (dy/dx) Δx

= 1/4(x)^{3/4} Δx [∵ y = (x)^{1/4}]

= 1/4(16)^{1/4} (- 1)

= - 1/(4 x 8)

= - 1/32

= - 0.03125

Hence,

(15)^{1/4}= 2 + (- 0.03125)

= 1.96875

Thus, the approximate value of (15)^{1/4 }= 1.96875.

(vii) (26)^{1/3}

Consider y = (x)^{1/3}

Let x = 27 and Δx = - 1

Then,

Δy = (x + Δx)^{1/3} - (x)^{1/3}

= (26)^{1/3} - (27)^{1/3}

= (26)^{1/3} - 3

Δy + 3 = (26)^{1/3}

Now, dy is approximately equal to Δy and is given by,

dy = (dy/dx) Δx

= 1/3(x)^{2/3} Δx [∵ y = (x)^{1/3}]

= 1/3(27)^{2/3} (- 1)

= - 0.0370

Hence,

(26)^{1/3 }= 3 + (- 0.0370)

= 2.9629

Thus, the approximate value of (26)^{1/3}= 2.9629.

(viii) (255)^{1/4}

Consider y = (x)^{1/4}

Let x = 256 and Δx = - 1

Then,

Δy = (x + Δx)^{1/4} - (x)^{1/4}

= (255)^{1/4} - (256)^{1/4}

= (255)^{1/3} - 4

Δy + 4 = (255)^{1/4}

Now, dy is approximately equal to Δy and is given by,

dy = (dy/dx) Δx

= 1/4(x)^{3/4} Δx [∵ y = (x)^{1/4}]

= 1/4(256)^{3/4} (- 1)

= - 1/4 x 4^{3}

= - 1/32

= - 0.0039

Hence,

(255)^{1/4 }= 4 + (- 0.0039)

= 3.9961

Thus, the approximate value of (255)^{1/4} = 3.9961.

(ix) (82)^{1/4}

Consider y = (x)^{1/4}

Let x = 81 and Δx = 1

Then,

Δy = (x + Δx)^{1/4} - (x)^{1/4}

= (82)^{1/4} - (81)^{1/4}

= (82)^{1/3} - 3

Δy + 3 = (82)^{1/4}

Now, dy is approximately equal to Δy and is given by,

dy = (dy/dx) Δx

= 1/4(x)^{3/4} Δx [∵ y = (x)^{1/4}]

= 1/4(81)^{3/4} (1)

= 1/4 x 3^{3}

= 1/108

= 0.009

Hence,

(82)^{1/4 }= 3 + 0.009

= 3.009

Thus, the approximate value of (82)^{1/4}^{ }= 3.009.

(x) (401)^{1/2}

Consider y = (x)^{1/2}

Let x = 400 and Δx = 1

Then,

Δy = (x + Δx)^{1/2} - (x)^{1/2}

= (401)^{1/2} - (400)^{1/2}

= (401)^{1/3} - 20

Δy + 20 = (401)^{1/2}

Now, dy is approximately equal to Δy and is given by,

dy = (dy/dx) Δx

= 1/2√x Δx [∵ y = (x)^{1/2}]

= 1/2 (20) (1)

= 1/40

= 0.025

Hence,

(401)^{1/2 }= 20 + 0.025

= 20.025

Thus, the approximate value of (401)^{1/2}^{ }= 20.025.

(xi) (0.0037)^{1/2}

Consider y = (x)^{1/2}

Let x = 0.0036 and Δx = 0.0001

Then,

Δy = (x + Δx)^{1/2} - (x)^{1/2}

= (0.0037)^{1/2} - (0.0036)^{1/2}

= (0.0037)^{1/2} - 0.06

Δy + 0.06 = (0.0037)^{1/2}

Now, dy is approximately equal to Δy and is given by,

dy = (dy/dx) Δx

= 1/2√x Δx [∵ y = (x)^{1/2}]

= 1/2 (0.006) (1)

= 0.0001/0.12

= 0.0008325

Hence,

(0.0037)^{1/2 }= 0.006 + 0.00083

= 0.06083

Thus, the approximate value of (0.0037)^{1/2} = 0.06083.

(xii) (26.57)^{1/3}

Consider y = (x)^{1/3}

Let x = 27 and Δx = - 0.43

Then,

Δy = (x + Δx)^{1/3} - (x)^{1/3}

= (26.57)^{1/3} - (27)^{1/3}

= (26.57)^{1/3} - 3

Δy + 3 = (26.57)^{1/3}

Now, dy is approximately equal to Δy and is given by,

dy = (dy/dx) Δx

= 1/3(x)^{2/3} Δx [∵ y = (x)^{1/3}]

= [1/(3 x 9)] (- 0.43)

= - 0.43/27

= - 0.015

Hence,

(26.57)^{1/3 }= 3 + (- 0.015)

= 2.984

Thus, the approximate value of (26.57)^{1/3 }= 2.984.

(xiii) (81.5)^{1/4}

Consider y = (x)^{1/4}

Let x = 81 and Δx = 0.5

Then,

Δy = (x + Δx)^{1/4} - (x)^{1/4}

= (81.5)^{1/4} - (81)^{1/4}

= (81.5)^{1/4} - 3

Δy + 3 = (81.5)^{1/4}

Now, dy is approximately equal to Δy and is given by,

dy = (dy/dx) Δx

= 1/4(x)^{3/4} Δx [∵ y = (x)^{1/4}]

= 1/4(81)^{3/4} (1)

= 1/4(3^{3}) (0.5)

= 0.5/108

= 0.0046

Hence,

(81.5)^{1/4 }= 3 + 0.0046

= 3.0046

Thus, the approximate value of (81.5)^{1/4}^{ }= 3.0046.

(xiv) (3.968)^{3/2}

Consider y = (x)^{3/2}

Let x = 4 and Δx = - 0.032

Then,

Δy = (x + Δx)^{3/2} - (x)^{3/2}

= (3.968)^{3/2} - (4)^{3/2}

= (3.968)^{3/2} - 8

Δy + 8 = (3.968)^{3/2}

Now, dy is approximately equal to Δy and is given by,

dy = (dy/dx) Δx

= 3/2(x)^{1/2} Δx [∵ y = (x)^{3/2}]

= 3/2 (2) (- 0.032)

= - 0.096

Hence,

(3.968)^{3/2 }= 8 + (- 0.096)

= 7.904

Thus, the approximate value of (3.968)^{3/2} = 7.904.

(xv) (32.15)^{1/5}

Consider y = (x)^{1/5}

Let x = 32 and Δx = 0.15

Then,

Δy = (x + Δx)^{1/5} - (x)^{1/5}

= (32.15)^{1/5} - (32)^{1/5}

= (32.15)^{1/5} - 2

Δy + 2 = (32.15)^{1/5}

Now, dy is approximately equal to Δy and is given by,

dy = (dy/dx) Δx

= 1/5(x)^{4/5} Δx [∵ y = (x)^{1/5}]

= 1/5(2)^{4/5} (0.15)

= 1/4(3^{3}) (0.5)

= 0.15/80

= 0.00187

Hence,

(32.15)^{1/5 }= 2 + 0.00187

= 2.00187

Thus, the approximate value of (32.15)^{1/5} = 2.00187

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.4 Question 1

## Using differentials, find the approximate value of each of the following up to 3 places of decimals.

**Summary:**

Using differentials, the approximate value of each of the following up to 3 places of decimals are 5.03, 7.035, 0.8, 0.208, 0.999, 1.968, 2.962, 3.996, 3.009, 20.025, 0.060,2.984, 3.004,7.904, 2.001 respectively

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