Approximating the Square Root of 0.036 using Derivatives

To approximate the square root of 0.036 using derivatives, we can use the concept of Taylor series expansion. The Taylor series expansion of a function around a point gives an approximation of the function using its derivatives.

Step 1: Define the Function
Let's define the function f(x) as the square root of x, i.e., f(x) = √x.

Step 2: Choose a Point for Expansion
Since we want to approximate the square root of 0.036, we can choose a point close to 0.036 for expansion. Let's choose x = 0.04 as the point for expansion.

Step 3: Calculate the Derivatives
To calculate the derivatives of f(x), we can differentiate the function with respect to x.

f'(x) = (1/2) * x^(-1/2) = 1 / (2√x)

Step 4: Taylor Series Expansion
The Taylor series expansion of a function f(x) around a point a is given by:

f(x) ≈ f(a) + f'(a)(x - a) + 1/2! * f''(a)(x - a)^2 + 1/3! * f'''(a)(x - a)^3 + ...

Using the first derivative, we can write the expansion of f(x) as:

f(x) ≈ f(a) + f'(a)(x - a)

Substituting the values:

f(x) ≈ √0.04 + (1 / (2√0.04))(x - 0.04)

Simplifying further:

f(x) ≈ 0.2 + 0.25(x - 0.04)

Step 5: Approximate the Square Root of 0.036
To approximate the square root of 0.036, we substitute x = 0.036 into the above equation:

√0.036 ≈ 0.2 + 0.25(0.036 - 0.04)

Calculating further:

√0.036 ≈ 0.2 + 0.25(-0.004)

√0.036 ≈ 0.2 - 0.001

√0.036 ≈ 0.199

Therefore, the approximate value of the square root of 0.036 using derivatives is 0.199.

By using derivatives and Taylor series expansion, we can approximate the value of a function at a given point. This method provides a useful tool for estimating mathematical expressions and can be applied to various functions and equations.