Differentials: Videos & Practice Problems

In calculus, understanding how to find derivatives is crucial, as it lays the foundation for exploring differentials. When we have a function represented as \( y = f(x) \), the derivative, denoted as \( f'(x) \), is the rate of change of \( y \) with respect to \( x \). Differentials, represented as \( dy \) and \( dx \), are closely related to this concept and can be derived from the derivative.

To find the differential \( dy \), we can use the relationship \( dy = f'(x) \cdot dx \). For example, consider the function \( f(x) = x^3 + x \). To find \( dy \) when \( x = 2 \) and \( dx = 0.1 \), we first need to compute the derivative \( f'(x) \). Using the power rule, we find:

\( f'(x) = 3x^2 + 1 \)

Substituting \( x = 2 \) into the derivative gives:

\( f'(2) = 3(2^2) + 1 = 12 + 1 = 13 \)

Now, we can calculate \( dy \) as follows:

\( dy = f'(2) \cdot dx = 13 \cdot 0.1 = 1.3 \)

This means that the differential \( dy \) is 1.3 when \( x = 2 \) and \( dx = 0.1 \).

Understanding the practical application of differentials is essential, especially in estimating values that are difficult to compute directly. For instance, to estimate \( f(2.1) \), we can use the relationship between \( dy \), \( dx \), and the function values. Here, \( dx = 0.1 \) and \( dy = 1.3 \) can be used to approximate:

\( f(2.1) \approx f(2) + dy \)

Calculating \( f(2) \) gives:

\( f(2) = 2^3 + 2 = 10 \)

Thus, we can estimate:

\( f(2.1) \approx 10 + 1.3 = 11.3 \)

This approximation shows how differentials can simplify complex calculations, providing a quick way to estimate function values near a known point. In real-world applications, such as scientific research, this method allows for efficient problem-solving when dealing with complicated equations.

To find the differential \( dy \) for the function \( f(x) = x^3 - 2x \) when \( x = 2 \) and \( dx = 0.2 \), we start by recalling that the differential \( dy \) is given by the formula:

\( dy = f'(x) \, dx \)

First, we need to compute the derivative \( f'(x) \). Using the power rule, we differentiate \( f(x) \):

\( f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(2x) = 3x^2 - 2 \)

Now, substituting \( x = 2 \) into the derivative:

\( f'(2) = 3(2^2) - 2 = 3(4) - 2 = 12 - 2 = 10 \)

Next, we can calculate \( dy \) by substituting \( f'(2) \) and \( dx \) into the differential formula:

\( dy = f'(2) \cdot dx = 10 \cdot 0.2 = 2 \)

Now that we have \( dy = 2 \), we can estimate \( f(2.2) \). The approximation can be expressed as:

\( f(x + dx) \approx f(x) + dy \)

In this case, we want to estimate \( f(2.2) \), which can be rewritten as:

\( f(2.2) \approx f(2) + dy \)

First, we calculate \( f(2) \):

\( f(2) = 2^3 - 2(2) = 8 - 4 = 4 \)

Now, substituting \( f(2) \) and \( dy \) into the approximation:

\( f(2.2) \approx 4 + 2 = 6 \)

Thus, the estimated value of \( f(2.2) \) is 6. This method of using differentials allows us to approximate function values effectively.

In mathematics, particularly in fields like engineering and science, approximating values is a common practice, often utilizing differentials. When we approximate a function, it’s crucial to assess how close our estimated value is to the actual or exact value. This evaluation is essential for ensuring the accuracy of our measurements.

One method to determine the accuracy of an approximation is through the concept of absolute error. Absolute error is defined as the difference between the exact value and the approximated value. Depending on the context, the exact value may also be referred to as the actual value. For instance, if we approximate a function and obtain a value of 11.3, we can find the exact value by substituting the relevant input into the function. If the exact calculation yields 11.361, we can calculate the absolute error using the formula:

Absolute Error = |Exact Value - Approximate Value|

Substituting our values gives us:

Absolute Error = |11.361 - 11.3| = 0.061

A smaller absolute error indicates a more accurate approximation. If the absolute error were significantly larger, it would suggest that the approximation method was not effective.

Another important measure is the relative error, which provides a ratio of the absolute error to the exact value. This is calculated using the formula:

Relative Error = Absolute Error / Exact Value

Using our previous example, the relative error would be:

Relative Error = 0.061 / 11.361 ≈ 0.005369

A smaller relative error indicates a more precise approximation. To express this relative error as a percentage, we multiply by 100%:

Percentage Error = Relative Error × 100%

Thus, in our case:

Percentage Error ≈ 0.005369 × 100% ≈ 0.5369%

This small percentage error signifies that our approximation was quite accurate. Understanding these concepts—absolute error, relative error, and percentage error—enables mathematicians, engineers, and scientists to evaluate the effectiveness of their approximation methods and ensure the reliability of their results.

To estimate the value of the function \( f(x) = 8x - x^2 \) at \( x = 3.1 \) using differentials, we start by recognizing that \( f(3.1) \) can be approximated as \( f(3) + dy \), where \( dy \) is the differential of \( f \). The differential \( dy \) is calculated using the formula:

\( dy = f'(x) \cdot dx \)

First, we need to find the derivative \( f'(x) \). Using the power rule, we differentiate \( f(x) \):

\( f'(x) = 8 - 2x \)

Next, we substitute \( x = 3 \) into the derivative:

\( f'(3) = 8 - 2(3) = 8 - 6 = 2 \)

Given that \( dx = 0.1 \), we can now calculate \( dy \):

\( dy = f'(3) \cdot dx = 2 \cdot 0.1 = 0.2 \)

Now, we can estimate \( f(3.1) \):

\( f(3.1) \approx f(3) + dy \)

To find \( f(3) \), we substitute \( x = 3 \) into the original function:

\( f(3) = 8(3) - 3^2 = 24 - 9 = 15 \)

Thus, we have:

\( f(3.1) \approx 15 + 0.2 = 15.2 \)

Next, we calculate the absolute error by finding the actual value of \( f(3.1) \):

\( f(3.1) = 8(3.1) - (3.1)^2 = 24.8 - 9.61 = 15.19 \)

The absolute error is given by:

\( \text{Absolute Error} = | \text{Actual Value} - \text{Approximation} | = | 15.19 - 15.2 | = 0.01 \)

To assess the accuracy of our approximation, we calculate the relative error, which is the ratio of the absolute error to the actual value:

\( \text{Relative Error} = \frac{\text{Absolute Error}}{\text{Actual Value}} = \frac{0.01}{15.19} \approx 0.0006583 \)

This small relative error indicates that our approximation was quite accurate. In summary, using differentials, we estimated \( f(3.1) \) to be approximately \( 15.2 \), with an absolute error of \( 0.01 \) and a relative error of approximately \( 0.0006583 \).