Problem 10 - Single Variable Calculus

Question

Find the derivative of the function. \(f(x)=0.3 x^{-1.2}\)

Step-by-Step Solution

Answer

The derivative of the function \(f(x) = 0.3x^{-1.2}\) is \(f'(x) = -0.36x^{-2.2}\).

Step 1: Identify the coefficients and powers of the function

In the given function, \(f(x) = 0.3x^{-1.2}\), we can see that the constant coefficient ‘a’ is 0.3 and the power ‘n’ is -1.2.

Step 2: Apply the power rule for differentiation

By applying the power rule for differentiation - \(f'(x) = anx^{n-1}\), we get: \(f'(x) = 0.3(-1.2)x^{-1.2 - 1}\)

Step 3: Simplify the equation

Next, we simplify the equation: \(f'(x) = -0.36x^{-2.2}\) So, the derivative of the function \(f(x) = 0.3x^{-1.2}\) is \(f'(x) = -0.36x^{-2.2}\).

Key Concepts

Power Rule

The power rule is a fundamental technique in calculus used to find the derivative of a function that is a power of a variable. Specifically, this rule states that for a function of the form \(f(x) = ax^n\), the derivative \(f'(x)\) is given by \(anx^{n-1}\). This simplifies the process of differentiation, allowing us to quickly find derivatives without having to rely on more complex methods.

• The coefficient \(a\) remains unchanged.
• The power \(n\) becomes the new coefficient.
• The new power of \(x\) is \(n-1\).

Let's see how this is applied in our example with the function \(f(x) = 0.3x^{-1.2}\):

• Identify \(a = 0.3\) and \(n = -1.2\).
• Multiply \(a\) by \(n\) to get the new coefficient: \(0.3 \times -1.2\).
• Subtract 1 from \(n\) to find the new power: \(-1.2 - 1\).

Thus, the derivative becomes \(f'(x) = -0.36x^{-2.2}\).

Derivatives

Derivatives are a key concept in calculus, representing the rate at which a function is changing at any given point. Essentially, the derivative of a function gives us a new function that provides the slope of the original function's graph at any point. This slope indicates how steeply the graph is ascending or descending.
Understanding derivatives allows us to explore a variety of applications, such as finding maximum and minimum values of functions, analyzing curves, and solving problems in physics and engineering.

• They measure how a function's output value changes as its input changes.
• Can be visualized as the tangent line to the function's graph at a point.
• Help in understanding the dynamics of complex systems by modeling the rate of changes.

For example, in the exercise, finding the derivative \(f'(x) = -0.36x^{-2.2}\) tells us exactly how \(f(x) = 0.3x^{-1.2}\) changes at any point \(x\).

Calculus

Calculus is a branch of mathematics focused on the concepts of change and motion. It is divided into two main branches: differential calculus and integral calculus. Differential calculus, which deals with derivatives, helps us understand how things change instantaneously.
Calculus forms the backbone of many scientific disciplines, offering tools to model and solve problems involving dynamic systems. It enables us to compute velocity, acceleration, and limits of a function—a crucial part of science and engineering problems.

• Involves limits, derivatives, integrals, and infinite series.
• Enables dealing with quantities that are continuously changing.
• Forms the language in which laws of physics are expressed.

By utilizing the power rule in calculus, as demonstrated in our example, we can quickly find how functions like \(f(x) = 0.3x^{-1.2}\) behave and change, stitching a clear path through the complex world of continuous change.