Question

Differentiate the following funetions: \(\pi x^{4}-3 \frac{8}{4} x^{3}+\sqrt{3}\)

Short answer

Question: Find the derivative of the given function: \(f(x) = \pi x^{4} - 3\frac{8}{4}x^{3} + \sqrt{3}\). Answer: The derivative of the function is \(f'(x) = 4\pi x^{3} - 6x^{2}\).

Step-by-step solution

Identify the terms of the function to be differentiated

The given function is: \(\pi x^{4} - 3\frac{8}{4}x^{3} + \sqrt{3}\). We can see that there are three terms in this function: 1. \(\pi x^{4}\) 2. \(-3\frac{8}{4}x^{3}\) 3. \(\sqrt{3}\) We will differentiate each term and then combine the results to find the derivative of the whole function.

Differentiate the first term (\(\pi x^{4}\)) using the power rule

According to the power rule, if we have a function \(f(x) = ax^n\), the derivative will be \(f'(x) = nax^{n-1}\). Applying this rule to the first term, we get: \(\frac{d}{dx} (\pi x^{4}) = 4\pi x^{3}\)

Differentiate the second term (-3\(\frac{8}{4}x^{3}\)) using the power rule

Applying the power rule to the second term, we get: \(\frac{d}{dx} (-3\frac{8}{4}x^{3}) = -3\frac{8}{4} \cdot 3x^{2} = -6x^{2}\)

Differentiate the third term (\(\sqrt{3}\))

The third term is a constant term, which means its derivative is zero: \(\frac{d}{dx} (\sqrt{3}) = 0\)

Combine the derivatives of all terms

Now, let's add the derivatives of all the terms to find the derivative of the whole function: \(\frac{d}{dx}(f(x)) = 4\pi x^{3} - 6x^{2} + 0\) Thus, the derivative of the function is: \(f'(x) = 4\pi x^{3} - 6x^{2}\)

Key concepts

Power Rule

The Power Rule is a fundamental technique in differentiation used to simplify the process of finding derivatives of polynomial functions. When using the power rule, you look at terms of the form \( ax^n \), where \( a \) is a constant and \( n \) is a positive integer. To differentiate such a term, you multiply the power \( n \) by the coefficient \( a \) and then reduce the power of \( x \) by one.

Here's the rule in LaTeX form for clarity:

• If \( f(x) = ax^n \), then its derivative is \( f'(x) = nax^{n-1} \).

Let's consider the example from the original exercise. The term \( \pi x^4 \) uses the power rule:

• Multiply \( 4 \) (the exponent) with \( \pi \) (the constant), resulting in \( 4\pi \).
• Decrease the exponent by one, giving \( x^{3} \).

So, the derivative of \( \pi x^4 \) is \( 4\pi x^3 \). This rule makes it easy to handle polynomial terms quickly and accurately.

Derivative of a Polynomial

Polyynomials are combinations of terms that include constants, variables raised to powers, and coefficients. Differentiating a polynomial involves finding the derivative of each term individually and then combining them. Here, the power rule will be frequently used.

Let's break down the example given:

• The polynomial function is \( \pi x^4 - 3\frac{8}{4}x^3 + \sqrt{3} \).
• Identify & separate each term to differentiate: \( \pi x^4 \), \( -3\frac{8}{4}x^3 \), and \( \sqrt{3} \).
• Use the power rule for \( \pi x^4 \) to get \( 4\pi x^3 \), and likewise for \( -6x^2 \) from \( -6x^3 \).

After finding the derivatives of each term as per the steps above, the final derivative of the entire polynomial is simply the sum of these individual derivatives. Thus, we derive the polynomial function to obtain \( f'(x) = 4\pi x^3 - 6x^2 \). Each term in the polynomial is considered, confirming the method's efficiency and simplicity.

Constant Rule

The Constant Rule is one of the simplest rules in calculus for differentiation. It states that the derivative of a constant value is always zero. This is because constants do not change, and differentiation measures how a function changes.

Consider the example given with the function \( \sqrt{3} \), where:

• A constant term like \( \sqrt{3} \) has no variable part \( x \).
• Thus, the slope or rate of change is \( 0 \), because it remains the same regardless of \( x \).
• Mathematically, \( \frac{d}{dx}(c) = 0 \) for any constant \( c \).

In the exercise, the constant \( \sqrt{3} \) ceases to contribute to the value of the derivative as it remains invariant. It's simple but can be easily overlooked, reinforcing the importance of understanding each component's role in differentiation.