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Find the first and second derivatives of the given function. \(f(x)=-0.2 x^{2}+0.3 x+4\)

Short Answer

Expert verified
The first derivative of the function is \(f'(x) = -0.4x + 0.3\), and the second derivative is \(f''(x) = -0.4\).

Step by step solution

01

Find the first derivative

To find the first derivative of the function \(f(x)= -0.2x^2 + 0.3x + 4\), we will apply the power rule, which states that if \(f(x) = ax^n\), then \(f'(x) = nax^{n-1}\). Differentiate each term in the function with respect to x: \(f'(x) = \frac{d}{dx}(-0.2x^2) + \frac{d}{dx}(0.3x) + \frac{d}{dx}(4)\) Applying the power rule, we get: \(f'(x) = -0.4x + 0.3\) So, the first derivative of the function is \(f'(x) = -0.4x + 0.3\).
02

Find the second derivative

Now we will find the second derivative of the function by differentiating the first derivative, \(f'(x) = -0.4x + 0.3\), with respect to x: \(f''(x) = \frac{d^2}{dx^2}(-0.4x + 0.3)\) Applying the power rule again, we get: \(f''(x) = -0.4\) So, the second derivative of the function is \(f''(x) = -0.4\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Derivative
Differential calculus often begins with the concept of the first derivative. It represents the rate at which a function is changing at any given point, and it's a fundamental tool for understanding behavior of functions, especially in physics and engineering. In practical terms, finding the first derivative involves applying rules of differentiation to the original function.

For instance, let's consider the function from the exercise, which is a simple polynomial. By utilizing the power rule, one of the most common rules of differentiation, we efficiently calculate the first derivative. If we have a closer look at our function, each term is differentiated individually. For a term like \(0.3x\), the first derivative would be just \(0.3\), as a constant multiplied by a variable to the first power simply results in the constant. For constant terms, such as \(4\), the derivative is \(0\), because a constant doesn't change and therefore, its rate of change is zero.
Second Derivative
After grasping the first derivative, the next step is the second derivative. This derivative of the derivative provides insights into the acceleration of the function, or how the rate of change itself is changing. It's crucial information when you're dealing with motion or trying to determine the concavity of a graph - that is, whether the graph is bending up or down.

In our exercise, we've found the first derivative to be a linear function, which still incorporates x, the variable. Now, when we differentiate \( -0.4x + 0.3 \) with respect to x, we're left with \( -0.4 \), a constant. This constant second derivative indicates that the original function has uniform concavity, and in this case, since the second derivative is negative, the function is concave down everywhere on its domain.
Power Rule
One of the bedrocks of differential calculus is the power rule, an efficient shortcut when differentiating polynomials. The power rule states that if you have a term in the form of \(ax^n\), its derivative with respect to x is given by \(nax^{n-1}\). This means you bring down the exponent as a multiplier to the coefficient and then subtract one from the exponent.

For example, when differentiating the term \( -0.2x^2 \) from our exercise, we multiply the exponent, 2, by the coefficient, -0.2, to get \( -0.4 \) as the new coefficient. We then decrease the exponent by one to get \( -0.4x \). This simplicity and elegance make the power rule a staple for students and professionals alike.
Differential Calculus
Differential calculus is a massive field in mathematics that deals with the study of rates at which quantities change. It's the tool we use to find derivatives and understand how a particular function behaves moment to moment. Think of it as a mathematical microscope that allows us to zoom in on an instant in the life of a function to see how it's moving right then and there.

Through differentiation, we can solve a variety of practical problems, from predicting the future state of a system to finding the optimal solution in a given scenario. Whether it's in the realm of economics to find maximum profit, or in physics to determine the velocity of an object, differential calculus is indispensable. The exercise we've discussed not only provides practice in computing derivatives but also offers a glimpse into the real-world applications of calculus.

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