Problem 53 - Calculus Single Variable

Question

In Exercises \(53-56,\) compute \(f^{\prime}(\mathrm{c})\) for the given \(f\) and \(c\). \(f(x)=x^{5}(2 x+1), c=-1\)

Step-by-Step Solution

Answer

The derivative \(f^{\prime}(-1)\) is \(-7\).

Step 1: Understand the Problem

We're asked to find the derivative of a given function \(f(x)\) evaluated at a specific point \(c\). The function given is \(f(x) = x^5(2x + 1)\) and \(c = -1\). This means we need to compute \(f'(x)\) and then substitute \(x = -1\) to find \(f'(-1)\).

Step 2: Simplify the Function

Before differentiating, it helps to simplify \(f(x)\). We expand the function: \[f(x) = x^5(2x + 1) = 2x^6 + x^5\] This expression makes it easier to apply basic differentiation rules.

Step 3: Differentiate the Function

Apply the power rule to differentiate each term of the simplified function. The power rule states that \(\frac{d}{dx}[x^n] = nx^{n-1}\):- Differentiate \(2x^6\): \(\frac{d}{dx}[2x^6] = 12x^5\)- Differentiate \(x^5\): \(\frac{d}{dx}[x^5] = 5x^4\)Thus, the derivative \(f'(x) = 12x^5 + 5x^4\).

Step 4: Evaluate the Derivative at \(c = -1\)

Substitute \(x = -1\) into the derivative \(f'(x)\): \[f'(-1) = 12(-1)^5 + 5(-1)^4\]Calculate each part:- \((-1)^5 = -1\), so \(12(-1)^5 = -12\) - \((-1)^4 = 1\), so \(5(-1)^4 = 5\)Add these results: \[f'(-1) = -12 + 5 = -7\]

Step 5: Conclusion

The derivative of the function \(f(x) = x^5(2x + 1)\) evaluated at \(c = -1\) is \(-7\).

Key Concepts

Power Rule in Differentiation

The power rule is a fundamental tool in calculus used to find the derivative of functions with power terms. Understanding this rule is crucial for efficiently computing derivatives. The power rule states that if you have a function in the form of any power of variable, like \( x^n \), the derivative is found using:

• \( \frac{d}{dx}[x^n] = nx^{n-1} \)

This rule essentially "brings down" the exponent as a coefficient and reduces the power by one.
For example, when differentiating \( x^5 \), applying the power rule gives us 5 as a coefficient and decreases the power to 4, resulting in \( 5x^4 \).
Using the power rule simplifies the process of differentiation, especially for polynomial functions, and helps find overall derivatives swiftly by applying it term by term.

Evaluating Derivatives at a Point

After we find a derivative, there are often cases where we need to evaluate it at a specific point, denoted by \( c \).
This process involves substituting the specific value into the derived equation to calculate how the function behaves at that point.
Let's take the derivative \( f'(x) = 12x^5 + 5x^4 \), derived in the given exercise. To evaluate this at \( x = -1 \), substitute -1 for every instance of \( x \) in the derivative:

• \( 12(-1)^5 = -12 \)
• \( 5(-1)^4 = 5 \)

Adding these results provides \( f'(-1) = -12 + 5 = -7 \).
This means, at \( x = -1 \), the slope or rate of change of the original function is -7. Evaluating derivatives at a point is crucial for understanding the behavior of functions at specific instances.

Polynomial Function Differentiation

Differentiating polynomial functions involves using the power rule separately on each term. A polynomial is composed of terms with varying powers of \( x \), such as \( ax^n \), where \( a \) is any constant, and \( n \) is a non-negative integer.
When tackling polynomials like the function \( f(x) = 2x^6 + x^5 \), the objective is to find the derivative of each term and sum them.
For \( 2x^6 \), the derivative is computed as \( 12x^5 \), and for \( x^5 \), it is \( 5x^4 \). Hence, the derivative of the polynomial is:

• \( f'(x) = 12x^5 + 5x^4 \)

Such differentiation is straightforward as it involves applying the power rule sequentially to achieve the derivative expression, making the process of finding slopes or rate of change at any point swift and manageable. Polynomials are foundational in calculus, and differentiating them is a critical skill for more advanced topics.