Find The Derivative Of The Function.$\[ U(y)=\left(\frac{y^4+5}{y^2+5}\right)^6 \\]$\[ U^{\prime}(y)=\square \\]

Introduction

In this article, we will explore the process of finding the derivative of a given function, U(y), which is a composite function involving a power and a quotient. The function is defined as:

$$U(y)=\left(\frac{y^4+5}{y^2+5}\right)^6$$

To find the derivative of U(y), we will apply the chain rule and the quotient rule of differentiation. The chain rule states that if we have a composite function of the form f(g(x)), then the derivative of f(g(x)) is given by f'(g(x)) * g'(x). The quotient rule states that if we have a function of the form f(x)/g(x), then the derivative of f(x)/g(x) is given by (f'(x) * g(x) - f(x) * g'(x)) / g(x)^2.

Step 1: Apply the Chain Rule

To find the derivative of U(y), we first apply the chain rule. Let's define the inner function as:

$$f(y)=\frac{y^4+5}{y^2+5}$$

Then, the derivative of f(y) is given by:

$$f^{\prime}(y)=\frac{(y^2+5)(4y^3)- (y^4+5)(2y)}{(y^2+5)^2}$$

Step 3: Simplify the Derivative of the Inner Function

We can simplify the derivative of the inner function by combining like terms:

$$f^{\prime}(y)=\frac{4y^5+20y-2y^5-10y}{(y^2+5)^2}$$

$$f^{\prime}(y)=\frac{2y^5+10y}{(y^2+5)^2}$$

Step 4: Apply the Chain Rule Again

Now, we can apply the chain rule again to find the derivative of U(y):

$$U^{\prime}(y)=6\left(\frac{y^4+5}{y^2+5}\right)^5 \cdot \frac{2y^5+10y}{(y^2+5)^2}$$

Step 5: Simplify the Derivative of U(y)

We can simplify the derivative of U(y) by combining like terms:

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \left(\frac{y^4+5}{y^2+5}\right)^5$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3<br/> # Find the Derivative of the Function U(y) - Q&A ## Introduction In our previous article, we explored the process of finding the derivative of a given function, U(y), which is a composite function involving a power and a quotient. The function is defined as: $U(y)=\left(\frac{y^4+5}{y^2+5}\right)^6

To find the derivative of U(y), we applied the chain rule and the quotient rule of differentiation. In this article, we will answer some frequently asked questions related to the derivative of U(y).

Q1: What is the derivative of U(y)?

A1: The derivative of U(y) is given by:

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

Q2: How do I apply the chain rule to find the derivative of U(y)?

A2: To apply the chain rule, we first define the inner function as:

$$f(y)=\frac{y^4+5}{y^2+5}$$

Then, we find the derivative of f(y) using the quotient rule:

$$f^{\prime}(y)=\frac{(y^2+5)(4y^3)- (y^4+5)(2y)}{(y^2+5)^2}$$

Next, we apply the chain rule again to find the derivative of U(y):

$$U^{\prime}(y)=6\left(\frac{y^4+5}{y^2+5}\right)^5 \cdot \frac{2y^5+10y}{(y^2+5)^2}$$

Q3: What is the quotient rule of differentiation?

A3: The quotient rule of differentiation states that if we have a function of the form f(x)/g(x), then the derivative of f(x)/g(x) is given by:

$$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{f^{\prime}(x) \cdot g(x) - f(x) \cdot g^{\prime}(x)}{g(x)^2}$$

Q4: How do I simplify the derivative of U(y)?

A4: To simplify the derivative of U(y), we can combine like terms and cancel out common factors. For example:

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

$$U^{\prime}(y)=\frac{12y^5+60y}{(y^2+5)^3} \cdot \frac{(y^4+5)^5}{(y^2+5)^5}$$

Q5: What is the significance of the derivative of U(y)?

A5: The derivative of U(y) represents the rate of change of the function U(y) with respect to the variable y. It can be used to analyze the behavior of the function and make predictions about its future values.

Q6: How do I use the derivative of U(y) in real-world applications?

A6: The derivative of U(y) can be used in a variety of real-world applications, such as:

• Modeling population growth and decline
• Analyzing the behavior of complex systems
• Making predictions about future values of a function
• Optimizing functions to minimize or maximize their values

Q7: What are some common mistakes to avoid when finding the derivative of a function?

A7: Some common mistakes to avoid when finding the derivative of a function include:

• Failing to apply the chain rule or quotient rule correctly
• Not simplifying the derivative properly
• Making errors in the calculation of the derivative
• Not checking the units of the derivative to ensure they are correct

Q8: How do I check the units of the derivative to ensure they are correct?

A8: To check the units of the derivative, we need to ensure that the units of the derivative match the units of the original function. For example, if the original function has units of meters per second, then the derivative should have units of meters per second squared.

Q9: What are some common applications of the derivative in physics?

A9: The derivative has many applications in physics, including:

• Modeling the motion of objects
• Analyzing the behavior of complex systems
• Making predictions about future values of a function
• Optimizing functions to minimize or maximize their values

Q10: How do I use the derivative to model the motion of an object?

A10: To use the derivative to model the motion of an object, we need to define the position function of the object and then find its derivative. The derivative of the position function represents the velocity of the object, and the derivative of the velocity function represents the acceleration of the object.

Conclusion

In this article, we have answered some frequently asked questions related to the derivative of U(y). We have also discussed the significance of the derivative and its applications in real-world scenarios. By understanding the derivative and its applications, we can gain a deeper understanding of the behavior of complex systems and make predictions about their future values.