Problem 27
Question
Use Example 6 as a model to find the derivative. $$ y=\frac{\sqrt{x}}{x} $$
Step-by-Step Solution
Verified Answer
The derivative of the function \(y=\frac{\sqrt{x}}{x}\) is \(y'=-\frac{1}{2\sqrt{x^3}}\).
1Step 1: Rewrite the Function
Transform the function into a form that's easier to differentiate. Rewrite the function as \(y=x^{-1/2}\). The square root in the numerator can be rewritten as \(x^{1/2}\) and, because it's in the denominator, it becomes \(x^{-1/2}\).
2Step 2: Use the Power Rule
The power rule for differentiation, \(\frac{d}{dx} x^n = n*x^{n-1}\), can be applied directly. Then, the derivative of the function \(y=x^{-1/2}\) is \(y'=-1/2*x^{-3/2}\).
3Step 3: Simplify the Derivative
Simplify the result by writing it in square root format. The final derivative is then \(y'=-\frac{1}{2\sqrt{x^3}}\).
Key Concepts
Power RuleDifferentiationSimplifying Derivatives
Power Rule
The Power Rule is a fundamental technique in calculus used to find the derivative of a function with a power.
This rule is particularly useful when dealing with polynomial functions or expressions that can be transformed into a power form. In simpler terms, the Power Rule states that if you have a term of the form \(x^n\), its derivative is determined by multiplying the power by the coefficient (if there is one) and reducing the power by one.
The formula can be expressed as follows:
This rule is particularly useful when dealing with polynomial functions or expressions that can be transformed into a power form. In simpler terms, the Power Rule states that if you have a term of the form \(x^n\), its derivative is determined by multiplying the power by the coefficient (if there is one) and reducing the power by one.
The formula can be expressed as follows:
- If \(y = x^n\), then the derivative \(y' = n \cdot x^{n-1}\).
Differentiation
Differentiation is the process of finding the derivative of a function.
It is a key concept in calculus and is used to determine the rate at which a function is changing at any given point. Differentiation can help us understand how quantities vary with one another and is the backbone of numerous applications in physics, engineering, and economics.
In our exercise, we needed to differentiate the function \(y=\frac{\sqrt{x}}{x}\). To do this, the function was first rewritten into a form more suitable for differentiation, that is, \(y=x^{-1/2}\). Once in the simpler form, using concepts like the Power Rule becomes straightforward. Differentiation essentially provides a procedure to analyze changes and behavior of functions efficiently and effectively.
It is a key concept in calculus and is used to determine the rate at which a function is changing at any given point. Differentiation can help us understand how quantities vary with one another and is the backbone of numerous applications in physics, engineering, and economics.
In our exercise, we needed to differentiate the function \(y=\frac{\sqrt{x}}{x}\). To do this, the function was first rewritten into a form more suitable for differentiation, that is, \(y=x^{-1/2}\). Once in the simpler form, using concepts like the Power Rule becomes straightforward. Differentiation essentially provides a procedure to analyze changes and behavior of functions efficiently and effectively.
Simplifying Derivatives
Simplifying derivatives is an essential step after finding the derivative of a function.
This process involves expressing the derivative in its simplest and most comprehensible form, often by converting complicated expressions back into more familiar terms. Simplification can help make results easier to understand and apply in further calculations.
For the function at hand, after applying the Power Rule, we obtained the derivative \(y' = -1/2 \cdot x^{-3/2}\). To simplify this, negative exponents were converted back into roots, such that \(x^{-3/2}\) was rewritten in square root form, resulting in \(y' = -\frac{1}{2\sqrt{x^3}}\). This form is much more intuitive and could be more readily used in practical applications or further calculus computations. Comprehending the significance of simplifying derivatives is key in achieving clarity and precision in your calculus work.
This process involves expressing the derivative in its simplest and most comprehensible form, often by converting complicated expressions back into more familiar terms. Simplification can help make results easier to understand and apply in further calculations.
For the function at hand, after applying the Power Rule, we obtained the derivative \(y' = -1/2 \cdot x^{-3/2}\). To simplify this, negative exponents were converted back into roots, such that \(x^{-3/2}\) was rewritten in square root form, resulting in \(y' = -\frac{1}{2\sqrt{x^3}}\). This form is much more intuitive and could be more readily used in practical applications or further calculus computations. Comprehending the significance of simplifying derivatives is key in achieving clarity and precision in your calculus work.
Other exercises in this chapter
Problem 26
Find the derivative of the function. State which differentiation rule(s) you used to find the derivative, $$ h(t)=\left(t^{5}-1\right)\left(4 t^{2}-7 t-3\right)
View solution Problem 27
Find the marginal profit for producing units. (The profit is measured in dollars.) $$ P=-2 x^{2}+72 x-145 $$
View solution Problem 27
Use the General Power Rule to find the derivative of the function. $$ h(x)=\left(6 x-x^{3}\right)^{2} $$
View solution Problem 27
find the given value. $$ f(x)=x^{2}\left(3 x^{2}+3 x-4\right) \quad f^{\prime \prime \prime}(-2) $$
View solution