Problem 55 - Calculus and its applications

Question

Find \(y^{\prime}\) $$ \text { If } y=\frac{2}{5 x^{3}}, \text { find }\left.\frac{d y}{d x}\right|_{x=4} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} \) is \(-\frac{3}{640}\) at \( x = 4 \).

Step 1: Identify the Problem

We need to find the derivative of the function \( y = \frac{2}{5x^3} \) with respect to \( x \), and then evaluate this derivative at \( x = 4 \).

Step 2: Rewrite the Function

Express the function in a form that is easier to differentiate. The given function can be rewritten as \( y = \frac{2}{5} x^{-3} \).

Step 3: Differentiate the Function

Apply the power rule for differentiation, \( \frac{d}{dx}(x^n) = n x^{n-1} \). For our function, \( y = \frac{2}{5} x^{-3} \), the derivative will be \( y' = \frac{2}{5} imes (-3) x^{-4} = -\frac{6}{5} x^{-4} \).

Step 4: Evaluate the Derivative at x = 4

Substitute \( x = 4 \) into the derivative: \( y' = -\frac{6}{5} (4)^{-4} = -\frac{6}{5} \times \frac{1}{256} = -\frac{6}{1280} = -\frac{3}{640} \).

Key Concepts

Understanding the Power Rule

The power rule is one of the fundamental tools we use in calculus to find the derivative of functions that are in the form of a variable raised to a power. This rule states that if you have a function in the form of \( x^n \), its derivative is given by \( \frac{d}{dx}(x^n) = n \times x^{n-1} \). Essentially, you bring the exponent in front as a coefficient and subtract one from the original exponent.

This rule applies straightforwardly to any power of \( x \), including negative and fractional powers. That's why in the original exercise, where the function \( y = \frac{2}{5} x^{-3} \), we used the power rule to derive \( y' = -\frac{6}{5} x^{-4} \). The exponent \(-3\) becomes \(-4\) after taking the derivative, following the power rule.

The power rule makes deriving polynomial functions much simpler and allows us to handle complex problems efficiently, with consistent outcomes for derivatives.

The Process of Function Differentiation

Differentiation is the process of calculating the derivative of a function, which essentially measures the rate at which a function is changing at any given point. It's a core concept in calculus, important for understanding how different functions behave.

In the context of the original exercise, differentiating the given function \( y = \frac{2}{5} x^{-3} \) involves rewriting it to a format more suitable for differentiation: \( y = \frac{2}{5} x^{-3} \). After this transformation, you can employ rules like the power rule with ease.

• Rewriting the function can simplify the differentiation.
• Apply rules consistently to ensure accurate outcomes.

By understanding how differentiation works, particularly in steps, learners can solidify their grasp on how changes in variables affect functions, which is crucial for more advanced math and physics.

Evaluating the Derivative

Once a function is differentiated, the next step often involves evaluating the derivative at specific points. This gives a numerical value that represents the rate of change of the function at that point.

In our exercise, after finding the derivative \( y' = -\frac{6}{5} x^{-4} \), we evaluated it at \( x = 4 \). Plugging in this value, we calculated \( y' = -\frac{6}{5} (4)^{-4} \), which simplifies to \( -\frac{3}{640} \).

This process helps in understanding not just how a function behaves but gives precise values needed for applications in science and engineering. Evaluating derivatives is an essential step in contextualizing abstract mathematical results to real-world scenarios.