Problem 6 - Calculus Single Variable

Question

Compute the indicated derivative for the given function by using the formulas and rules that are summarized at the end of this section. $$ D(H)(1), H(x)=10 / x $$

Step-by-Step Solution

Answer

The indicated derivative \( D(H)(1) \) is \(-10\).

Step 1: Identify the Function and the Task

The function given is \( H(x) = \frac{10}{x} \). The task is to find the derivative \( D(H)(1) \), which means we need to find \( H'(x) \) first, and then evaluate it at \( x = 1 \).

Step 2: Rewrite the Function

To make differentiating easier, let's rewrite the function. The function \( H(x) = \frac{10}{x} \) can be expressed as \( H(x) = 10x^{-1} \) for easier differentiation.

Step 3: Differentiate Using the Power Rule

Apply the power rule, which states that the derivative of \( ax^n \) is \( n \cdot ax^{n-1} \). For \( H(x) = 10x^{-1} \), \( a = 10 \) and \( n = -1 \). Thus, \( H'(x) = -1 \cdot 10x^{-2} = -10x^{-2} \).

Step 4: Simplify the Derivative Expression

The derivative \( H'(x) = -10x^{-2} \) can be rewritten as \( H'(x) = -\frac{10}{x^2} \) for better clarity.

Step 5: Evaluate the Derivative at \( x = 1 \)

Now substitute \( x = 1 \) into \( H'(x) = -\frac{10}{x^2} \). Thus, \( H'(1) = -\frac{10}{1^2} = -10 \).

Key Concepts

Power Rule

Learning the power rule is essential for students delving into the world of calculus. It makes finding derivatives a lot simpler and quicker. The power rule states that for any function of the form \( ax^n \), the derivative is given by \( n \cdot ax^{n-1} \). This means you take the exponent, multiply it by the coefficient of \( x \), and then reduce the exponent by one.

For example, if \( H(x) = 10x^{-1} \), where \( a = 10 \) and \( n = -1 \), applying the power rule yields \( H'(x) = -1 \cdot 10x^{-2} \). This results in \( H'(x) = -10x^{-2} \).

• It's a straightforward formula: \( n \cdot ax^{n-1} \).
• Allowing rapid differentiation of polynomial functions.
• Key for simplifying complex problems.

By understanding and applying the power rule correctly, calculus becomes a less daunting and more accurate venture.

Differentiation Techniques

Understanding differentiation techniques enhances your ability to tackle a wide range of calculus problems. Differentiation is essentially a way to find out how a function changes at any point.

There are many rules and techniques to help with this, like the sum rule, product rule, quotient rule, and chain rule, but the power rule is often your best friend for simplifying problems with polynomials. In this context, our focus is on recognizing when to use these strategies and how they interplay.

• Quotient Rule: Useful when dealing with divisors, if not rewriting terms.
• Rewriting Technique: Change division into negative exponents (turning \( \frac{10}{x} \) into \( 10x^{-1} \)).
• Bespoke Simplifications: Can convert a complex expression to a simple one, like turning \( x^{-2} \) into \( \frac{1}{x^2} \).

Mastering these techniques allows for seamless problem-solving and boosts confidence in handling calculus challenges.

Function Evaluation

Once you've found a derivative, evaluating it at a particular point is often required. This means substituting a value into the derivative to find the rate of change at that specific point.

From our example, after the power rule application gives \( H'(x) = -\frac{10}{x^2} \), evaluating the derivative at \( x = 1 \) involves plugging 1 into this expression. You find \( H'(1) = -\frac{10}{1^2} \), which simplifies to \( -10 \).

• It's a straightforward process: substitute \( x \) in the derivative with the given value.
• Direct evaluation gives the instantaneous rate of change, essential to many real-world applications.
• It also helps confirm the understanding of problems and their solutions.

With these skills, evaluating derivatives becomes a powerful tool for exploring dynamics and changes in functions across many fields.