Assignment #1 Due MondayDifferentiate The Following:i. $\[ Y = 5a^4 - 4a^3 + A^2 - 3 \\]ii. $\[ Y = X^3 + X \\]iii. $\[ Y = (2 - 6x)^8 \\]iv. $\[ Y = (3 - 2x)(2 - 4x) \\]v. $\[ Y = (x^2 + 2x + 1)(x^3 - X - 12)

Introduction

In this assignment, we will be differentiating five different functions. Differentiation is a fundamental concept in calculus that involves finding the derivative of a function, which represents the rate of change of the function with respect to one of its variables. In this assignment, we will apply the rules of differentiation to each of the given functions and find their derivatives.

Function i: $y = 5a^4 - 4a^3 + a^2 - 3$

The first function we will differentiate is $y = 5a^4 - 4a^3 + a^2 - 3$. To find the derivative of this function, we will apply the power rule of differentiation, which states that if $y = x^n$, then $y' = nx^{n-1}$.

Using the power rule, we can differentiate each term in the function separately:

• The derivative of $5a^4$ is $20a^3$.
• The derivative of $-4a^3$ is $-12a^2$.
• The derivative of $a^2$ is $2a$.
• The derivative of $-3$ is $0$, since the derivative of a constant is always $0$.

Therefore, the derivative of the function $y = 5a^4 - 4a^3 + a^2 - 3$ is:

$$y' = 20a^3 - 12a^2 + 2a$$

Function ii: $y = x^3 + x$

The second function we will differentiate is $y = x^3 + x$. To find the derivative of this function, we will again apply the power rule of differentiation.

Using the power rule, we can differentiate each term in the function separately:

• The derivative of $x^3$ is $3x^2$.
• The derivative of $x$ is $1$.

Therefore, the derivative of the function $y = x^3 + x$ is:

$$y' = 3x^2 + 1$$

Function iii: $y = (2 - 6x)^8$

The third function we will differentiate is $y = (2 - 6x)^8$. To find the derivative of this function, we will apply the chain rule of differentiation, which states that if $y = f(g(x))$, then $y' = f'(g(x)) \cdot g'(x)$.

In this case, we have $y = (2 - 6x)^8$, so we can let $f(u) = u^8$ and $g(x) = 2 - 6x$. Then, we have:

• $f'(u) = 8u^7$
• $g'(x) = -6$

Using the chain rule, we can find the derivative of the function $y = (2 - 6x)^8$ as follows:

$$y' = f'(g(x)) \cdot g'(x) = 8(2 - 6x)^7 \cdot (-6) = -48(2 - 6x)^7$$

Function iv: $y = (3 - 2x)(2 - 4x)$

The fourth function we will differentiate is $y = (3 - 2x)(2 - 4x)$. To find the derivative of this function, we will apply the product rule of differentiation, which states that if $y = u \cdot v$, then $y' = u' \cdot v + u \cdot v'$.

In this case, we have $y = (3 - 2x)(2 - 4x)$, so we can let $u = 3 - 2x$ and $v = 2 - 4x$. Then, we have:

• $u' = -2$
• $v' = -4$

Using the product rule, we can find the derivative of the function $y = (3 - 2x)(2 - 4x)$ as follows:

$$y' = u' \cdot v + u \cdot v' = (-2)(2 - 4x) + (3 - 2x)(-4) = -4 + 8x - 12 + 8x = -16 + 16x$$

Function v: $y = (x^2 + 2x + 1)(x^3 - x - 12)$

The fifth and final function we will differentiate is $y = (x^2 + 2x + 1)(x^3 - x - 12)$. To find the derivative of this function, we will again apply the product rule of differentiation.

Using the product rule, we can find the derivative of the function $y = (x^2 + 2x + 1)(x^3 - x - 12)$ as follows:

• Let $u = x^2 + 2x + 1$ and $v = x^3 - x - 12$.
• Then, we have $u' = 2x + 2$ and $v' = 3x^2 - 1$.
• Using the product rule, we can find the derivative of the function $y = (x^2 + 2x + 1)(x^3 - x - 12)$ as follows:

$$y' = u' \cdot v + u \cdot v' = (2x + 2)(x^3 - x - 12) + (x^2 + 2x + 1)(3x^2 - 1)$$

Simplifying the expression, we get:

$$y' = 2x^4 - 2x^3 - 24x + 2x^3 - 2x - 24 + 3x^4 - x^2 + 2x^3 - x = 5x^4 + x^3 - x^2 - 25x - 24$$

Conclusion

In this assignment, we have differentiated five different functions using the power rule, chain rule, and product rule of differentiation. We have found the derivatives of each function and simplified the resulting expressions. The derivatives of the functions are:

• $y' = 20a^3 - 12a^2 + 2a$
• $y' = 3x^2 + 1$
• $y' = -48(2 - 6x)^7$
• $y' = -16 + 16x$
• $y' = 5x^4 + x^3 - x^2 - 25x - 24$

Introduction

In this article, we will continue to discuss the differentiation of various functions. We will provide a Q&A section to help clarify any doubts or questions that readers may have.

Q&A

Q: What is the power rule of differentiation?

A: The power rule of differentiation is a fundamental rule in calculus that states that if $y = x^n$, then $y' = nx^{n-1}$. This rule can be used to differentiate any function that can be written in the form $y = x^n$.

Q: How do I apply the power rule of differentiation?

A: To apply the power rule of differentiation, you need to identify the exponent of the variable in the function. Then, you multiply the exponent by the coefficient of the variable and subtract 1 from the exponent. The result is the derivative of the function.

Q: What is the chain rule of differentiation?

A: The chain rule of differentiation is a rule that is used to differentiate composite functions. A composite function is a function that is written in the form $y = f(g(x))$. The chain rule states that if $y = f(g(x))$, then $y' = f'(g(x)) \cdot g'(x)$.

Q: How do I apply the chain rule of differentiation?

A: To apply the chain rule of differentiation, you need to identify the outer function and the inner function in the composite function. Then, you find the derivative of the outer function with respect to the inner function and multiply it by the derivative of the inner function.

Q: What is the product rule of differentiation?

A: The product rule of differentiation is a rule that is used to differentiate the product of two functions. The product rule states that if $y = u \cdot v$, then $y' = u' \cdot v + u \cdot v'$.

Q: How do I apply the product rule of differentiation?

A: To apply the product rule of differentiation, you need to identify the two functions that are being multiplied together. Then, you find the derivative of each function separately and multiply them together.

Q: What are some common mistakes to avoid when differentiating functions?

A: Some common mistakes to avoid when differentiating functions include:

• Forgetting to apply the power rule of differentiation when differentiating a function with a variable in the exponent.
• Forgetting to apply the chain rule of differentiation when differentiating a composite function.
• Forgetting to apply the product rule of differentiation when differentiating the product of two functions.
• Not simplifying the resulting expression after differentiating a function.

Conclusion

In this article, we have provided a Q&A section to help clarify any doubts or questions that readers may have about differentiating functions. We have discussed the power rule, chain rule, and product rule of differentiation, and provided examples of how to apply each rule. We have also identified some common mistakes to avoid when differentiating functions.

Practice Problems

To practice differentiating functions, try the following problems:

1. Differentiate the function $y = x^4 + 2x^3 - 3x^2 + x - 1$.
2. Differentiate the function $y = (2x + 1)^5$.
3. Differentiate the function $y = (x^2 + 1)(x^3 - x + 1)$.
4. Differentiate the function $y = \frac{x^2 + 1}{x + 1}$.
5. Differentiate the function $y = \sin(x^2 + 1)$.

Answer Key

1. $y' = 4x^3 + 6x^2 - 6x + 1$
2. $y' = 10(2x + 1)^4 \cdot 2$
3. $y' = (2x + 1)^5 \cdot (x^2 + 1) + (x^2 + 1)(3x^2 - 1)$
4. $y' = \frac{(x + 1)(2x) - (x^2 + 1)(1)}{(x + 1)^2}$
5. $y' = 2x \cos(x^2 + 1)$

We hope this article has been helpful in clarifying any doubts or questions that readers may have about differentiating functions.