Question Statement

Use the concept of the differential to find an approximation for $(1.8)^{5}$ .

Background and Explanation

Differentials provide a linear approximation of how a function changes with small changes in its input. For a function $y = f (x)$ , the differential is defined as $d y = f^{'} (x) d x$ , representing the principal (linear) part of the increment $Δ y$ . This concept extends to approximating function values near known points using the formula $f (x + Δ x) \approx f (x) + f^{'} (x) Δ x$ .

Solution

To find the approximation of $(1.8)^{5}$ , we follow these steps:

1. Identify the function and the point of interest We want to approximate $(1.8)^{5}$ , so we define the function: $f (x) = x^{5}$

We rewrite $1.8$ as a value near a known integer power, which is $2 - 0.2$ : $(1.8)^{5} = (2 - 0.2)^{5}$

2. Define the base value and the change We choose:

$x = 2$ (the base value where $2^{5}$ is easy to calculate)
$Δ x = d x = - 0.2$ (the small change from the base value)

3. Differentiate the function Differentiate $f (x)$ with respect to $x$ using the power rule: $f^{'} (x) = 5 x^{5 - 1} = 5 x^{4}$

4. Apply the linear approximation formula Using the formula $f (x + Δ x) \approx f (x) + f^{'} (x) d x$ , we substitute our values: $(x + Δ x)^{5} (2 + (- 0.2))^{5} (1.8)^{5} \approx x^{5} + 5 x^{4} d x \approx (2)^{5} + 5 (2)^{4} (- 0.2) \approx 32 + 5 (16) (- 0.2) \approx 32 - 16 = 16$

Hence, the approximation for $(1.8)^{5}$ is $16$ .

Key Formulas or Methods Used

Definition of Differential: $d y = f^{'} (x) d x$
Increment Method: $Δ y = f (x + Δ x) - f (x)$
Trigonometric Identity: $sin A - sin B = 2 cos (\frac{A + B}{2}) sin (\frac{A - B}{2})$
Linear Approximation Formula: $f (x + Δ x) \approx f (x) + f^{'} (x) Δ x$
Power Rule for Differentiation: $\frac{d}{d x} (x^{n}) = n x^{n - 1}$

Summary of Steps

Set $f (x) = x^{5}$ with a base point $x = 2$ and a change $Δ x = - 0.2$ .
Compute the function value at the base point: $f (2) = 2^{5} = 32$ .
Compute the derivative value at the base point: $f^{'} (2) = 5 (2)^{4} = 80$ .
Apply the approximation formula: $f (1.8) \approx 32 + 80 (- 0.2) = 16$ .

Background and Explanation

Differentials provide a linear approximation of how a function changes with small changes in its input. For a function

y = f (x)

, the differential is defined as

d y = f^{'} (x) d x

, representing the principal (linear) part of the increment

Δ y

. This concept extends to approximating function values near known points using the formula

f (x + Δ x) \approx f (x) + f^{'} (x) Δ x

Solution

To find the approximation of

(1.8)^{5}

, we follow these steps:

1. Identify the function and the point of interest We want to approximate

(1.8)^{5}

, so we define the function:

f (x) = x^{5}

We rewrite

1.8

as a value near a known integer power, which is

2 - 0.2

(1.8)^{5} = (2 - 0.2)^{5}

2. Define the base value and the change We choose:

x = 2

(the base value where

2^{5}

is easy to calculate)

Δ x = d x = - 0.2

(the small change from the base value)

3. Differentiate the function Differentiate

f (x)

with respect to

x

using the power rule:

f^{'} (x) = 5 x^{5 - 1} = 5 x^{4}

4. Apply the linear approximation formula Using the formula

f (x + Δ x) \approx f (x) + f^{'} (x) d x

, we substitute our values:

(x + Δ x)^{5} (2 + (- 0.2))^{5} (1.8)^{5} \approx x^{5} + 5 x^{4} d x \approx (2)^{5} + 5 (2)^{4} (- 0.2) \approx 32 + 5 (16) (- 0.2) \approx 32 - 16 = 16

Hence, the approximation for

(1.8)^{5}

16

2.7 Q-29

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps

2.7 Q-29

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps