Problem 47
Question
For the given \(f(x)\), find a formula for \(f^{\prime}(a)\). $$f(x)=\sqrt{x}$$
Step-by-Step Solution
Verified Answer
\(f'(a) = \frac{1}{2\sqrt{a}}\)
1Step 1: Understanding the function
We are given the function \(f(x) = \sqrt{x}\). This can also be expressed as \(f(x) = x^{1/2}\). We need to find the derivative \(f'(x)\) and then evaluate it at \(x = a\).
2Step 2: Using the Power Rule for Derivatives
The Power Rule states that the derivative of \(x^n\) is \(nx^{n-1}\). Here, \(n = 1/2\). Thus, the derivative is: \[f'(x) = \frac{1}{2}x^{-1/2}.\]
3Step 3: Substitute a into f'(x)
Now, substitute \(a\) into the derivative found in the previous step to find \(f'(a)\). This gives: \[f'(a) = \frac{1}{2}a^{-1/2}.\]
4Step 4: Simplify the Expression
Rewrite \(a^{-1/2}\) as \(\frac{1}{\sqrt{a}}\). Therefore, the expression for \(f'(a)\) simplifies to: \[f'(a) = \frac{1}{2\sqrt{a}}.\]
Key Concepts
Power RuleDifferentiationSimplifying Expressions
Power Rule
The power rule is a simple but fundamental concept in calculus used to find the derivative of functions in the form of a power, such as \(x^n\). According to the power rule, the derivative of \(x^n\) is \(nx^{n-1}\). This means you multiply by the current exponent, then subtract one from it to get the new exponent.
Applying the power rule makes differentiation much quicker, especially when dealing with functions like polynomial expressions.
In our original exercise, where we have \(f(x) = \sqrt{x}\), this can be rewritten as \(f(x) = x^{1/2}\). By applying the power rule, where \(n = \frac{1}{2}\), we find the derivative \(f'(x) = \frac{1}{2}x^{-1/2}\).
You see how we simply transferred our understanding of \(n\) to get the solution, which makes the power rule a go-to strategy in calculus.
Applying the power rule makes differentiation much quicker, especially when dealing with functions like polynomial expressions.
In our original exercise, where we have \(f(x) = \sqrt{x}\), this can be rewritten as \(f(x) = x^{1/2}\). By applying the power rule, where \(n = \frac{1}{2}\), we find the derivative \(f'(x) = \frac{1}{2}x^{-1/2}\).
You see how we simply transferred our understanding of \(n\) to get the solution, which makes the power rule a go-to strategy in calculus.
Differentiation
Differentiation is the process of finding the derivative of a function. The derivative measures how a function changes as its input changes, which is the essence of calculus.
In simpler terms, it represents the rate of change or the slope of the function at any given point.
When we differentiate functions, one of our core goals is to determine this derivative, which then helps us understand the function's behavior better, like finding increasing or decreasing intervals.
In our example, to differentiate \(f(x) = \sqrt{x}\), we translate this expression to \(x^{1/2}\) and apply the power rule to find \(f'(x)\). This derivative tells us how \(\sqrt{x}\) changes at any point, \(x\).
It’s crucial because derivatives form the backbone for more complex calculus concepts like integration and finding critical points.
In simpler terms, it represents the rate of change or the slope of the function at any given point.
When we differentiate functions, one of our core goals is to determine this derivative, which then helps us understand the function's behavior better, like finding increasing or decreasing intervals.
In our example, to differentiate \(f(x) = \sqrt{x}\), we translate this expression to \(x^{1/2}\) and apply the power rule to find \(f'(x)\). This derivative tells us how \(\sqrt{x}\) changes at any point, \(x\).
It’s crucial because derivatives form the backbone for more complex calculus concepts like integration and finding critical points.
Simplifying Expressions
Simplifying expressions is essential to make derivative results more interpretable and often easier to work with.
Simplifying involves rewriting expressions in their simplest form without altering their values, which often involves reducing a fraction or expressing a negative exponent differently.
Regarding our task, once we find the derivative \(f'(x) = \frac{1}{2}x^{-1/2}\), substituting \(x = a\) gives us \(f'(a) = \frac{1}{2}a^{-1/2}\).
We can make this expression cleaner by rewriting \(a^{-1/2}\) as \(\frac{1}{\sqrt{a}}\). Thus, \(f'(a) = \frac{1}{2\sqrt{a}}\).
Simplifying involves rewriting expressions in their simplest form without altering their values, which often involves reducing a fraction or expressing a negative exponent differently.
Regarding our task, once we find the derivative \(f'(x) = \frac{1}{2}x^{-1/2}\), substituting \(x = a\) gives us \(f'(a) = \frac{1}{2}a^{-1/2}\).
We can make this expression cleaner by rewriting \(a^{-1/2}\) as \(\frac{1}{\sqrt{a}}\). Thus, \(f'(a) = \frac{1}{2\sqrt{a}}\).
- This step is invaluable because it prepares the derivative for easier application in related problems.
- Simplified expressions like this are less intimidating and more directly useful in many mathematical contexts.
Other exercises in this chapter
Problem 46
Determine each limit. $$\lim _{x \rightarrow \infty} \frac{3 x^{3}+2 x-1}{2 x^{4}-3 x^{3}-2}$$
View solution Problem 47
Use a table and/or graph to decide whether each limit exists. If a limit exists, find its value. \(\lim _{x \rightarrow 1} f(x),\) where \(f(x)=\left\\{\begin{a
View solution Problem 47
Determine each limit, if it exists. $$\lim _{x \rightarrow 0} \frac{\sin x-3 x}{x}$$
View solution Problem 47
Determine each limit. $$\lim _{x \rightarrow \infty} \frac{2 x^{2}-1}{3 x^{4}+2}$$
View solution