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For each function, find the value of the derivative \(f^{\prime}(a)\) for the given value of \(a\) a. \(f(x)=x^{2}, a=1\) b. \(f(x)=x^{2}+3 x+1, a=3\) c. \(f(x)=\sqrt{x+1}, a=0\) d. \(f(x)=\frac{5}{x}, a=-1\)

Short Answer

Expert verified
a. 2, b. 9, c. 0.5, d. -5

Step by step solution

01

Differentiate the first function

The first function is given by \(f(x) = x^2\). To find the derivative, we use the power rule: \(f'(x) = 2x^{2-1} = 2x\).
02

Evaluate at the given point for (a)

We need to find \(f'(1)\) for the function \(f(x) = x^2\). Substitute \(x = 1\) into \(f'(x) = 2x\). This gives us \(f'(1) = 2(1) = 2\).
03

Differentiate the second function

The second function is \(f(x) = x^2 + 3x + 1\). The derivative is found by differentiating each term separately: \(f'(x) = 2x + 3\).
04

Evaluate at the given point for (b)

We need \(f'(3)\) for the function \(f(x) = x^2 + 3x + 1\). Substitute \(x = 3\) into \(f'(x) = 2x + 3\). This gives us \(f'(3) = 2(3) + 3 = 6 + 3 = 9\).
05

Differentiate the third function

The third function is \(f(x) = \sqrt{x+1}\). Using the chain rule, rewrite it as \((x+1)^{1/2}\) to find \(f'(x) = \frac{1}{2}(x+1)^{-1/2}\).
06

Evaluate at the given point for (c)

We need \(f'(0)\) for \(f(x) = \sqrt{x+1}\). Substitute \(x = 0\) into \(f'(x) = \frac{1}{2}(x+1)^{-1/2}\). This gives \(f'(0) = \frac{1}{2}(0+1)^{-1/2} = \frac{1}{2}(1)^{-1/2} = \frac{1}{2}\).
07

Differentiate the fourth function

The fourth function is \(f(x) = \frac{5}{x}\). Its derivative is found using the quotient rule or rewriting as \(5x^{-1}\): \(f'(x) = -5x^{-2}\) or \(-\frac{5}{x^2}\).
08

Evaluate at the given point for (d)

We need \(f'(-1)\) for \(f(x) = \frac{5}{x}\). Substitute \(x = -1\) into \(f'(x) = -\frac{5}{x^2}\). This gives \(f'(-1) = -\frac{5}{(-1)^2} = -\frac{5}{1} = -5\).

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Key Concepts

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

power rule
The power rule is a fundamental concept in calculus for differentiating functions of the form \( f(x) = x^n \). It provides a simple formula to compute derivatives quickly: if \( n \) is any real number, then the derivative of \( x^n \) is given by \( f'(x) = nx^{n-1} \). This rule is quite handy and transforms differentiation into a straightforward process.
When you apply the power rule, you reduce the power of \( x \) by one and multiply the expression by the original power, \( n \).
For example:
  • For \( f(x) = x^2 \), using the power rule gives \( f'(x) = 2x^{2-1} = 2x \).

This rule simplifies the differentiation of polynomial functions significantly. Whenever you have terms like \( x^2, x^3, \) or even \( x^{1/2} \), the power rule is your friend. Remember that when \( n=1 \), as in the term \( 3x \) found in the function \( f(x) = x^2 + 3x + 1 \), the derivative is simply \( 3 \), because \( 1 imes x^{0} = 1 \). Differentiation becomes a quick and powerful tool when employing the power rule.
chain rule
The chain rule is a technique used to differentiate compositions of functions. Essentially, if a function is composed of two or more functions, the chain rule helps find its derivative. It is particularly useful when dealing with nested functions or functions raised to a power.
The general idea is that you differentiate the outer function and then multiply by the derivative of the inner function.
For instance, consider the function \( f(x) = \sqrt{x+1} \).
  • Initially, we rewrite \( f(x) \) as \( (x+1)^{1/2} \).
  • The outer function is \( u^{1/2} \) and the inner function is \( u = x+1 \).
  • According to the chain rule, the derivative \( f'(x) \) is \( \frac{1}{2}(x+1)^{-1/2} \cdot 1 \), where \( 1 \) is the derivative of \( x+1 \).

This gives \( f'(x) = \frac{1}{2}(x+1)^{-1/2} \), which simplifies to \( \frac{1}{2\sqrt{x+1}} \). The chain rule comes into play whenever functions are dependent on other functions and serve to simplify the differentiation process.
quotient rule
The quotient rule is a method used to differentiate functions that are ratios of two functions. Whenever you see a function expressed as a fraction \( \frac{u(x)}{v(x)} \), the quotient rule becomes crucial to find its derivative.
According to this rule, if \( f(x) = \frac{u(x)}{v(x)} \), then the derivative \( f'(x) \) is computed as:
  • \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \)

This formula can look a bit complex at first, but it's quite manageable with practice. For the function \( f(x) = \frac{5}{x} \), you can look at it like \( u(x) = 5 \) and \( v(x) = x \).
  • The derivative \( u'(x) \) is 0 because 5 is a constant.
  • The derivative \( v'(x) \) is 1.

The derivative of the function following the quotient rule is \( f'(x) = \frac{0 \cdot x - 5 \cdot 1}{x^2} \), which simplifies to \( -\frac{5}{x^2} \).
Alternatively, by recognizing \( \frac{5}{x} \) as \( 5x^{-1} \), the derivative can also be found using the power rule. This flexibility in methods often simplifies problem-solving.

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