Find The Derivatives Of The Following Functions. Show All Steps For Full Credit.(a) Y = Sec ⁡ 101 ( 5 X Y = \sec^{101}(5x Y = Sec 101 ( 5 X ](b) H ( X ) = 10 Cos ⁡ ( 5 X ) 17 H(x) = \frac{10 \cos(5x)}{17} H ( X ) = 17 10 C O S ( 5 X ) ​ (c) Y = Ln ⁡ ( 12 X 3 − 5 X + 7 ) 21 Y = \ln\left(12x^3 - 5x + 7\right)^{21} Y = Ln ( 12 X 3 − 5 X + 7 ) 21 (d) $g(x) = 200x^8 \cdot

In this article, we will explore the derivatives of various functions, including trigonometric, logarithmic, and polynomial functions. We will use the chain rule, product rule, and quotient rule to find the derivatives of these functions.

(a) Derivative of $y = \sec^{101}(5x)$

To find the derivative of $y = \sec^{101}(5x)$, we will use the chain rule. The chain rule states that if we have a composite function of the form $y = f(g(x))$, then the derivative of $y$ with respect to $x$ is given by $y' = f'(g(x)) \cdot g'(x)$.

In this case, we have $y = \sec^{101}(5x)$, which can be written as $y = (\sec(5x))^{101}$. We can then apply the chain rule to find the derivative of $y$.

Step 1: Find the derivative of the outer function, which is $(\sec(5x))^{101}$.

Using the power rule, we have $\frac{d}{dx}((\sec(5x))^{101}) = 101(\sec(5x))^{100} \cdot \frac{d}{dx}(\sec(5x))$.

Step 2: Find the derivative of the inner function, which is $\sec(5x)$.

Using the chain rule, we have $\frac{d}{dx}(\sec(5x)) = \sec(5x) \tan(5x) \cdot \frac{d}{dx}(5x)$.

Step 3: Find the derivative of the innermost function, which is $5x$.

Using the power rule, we have $\frac{d}{dx}(5x) = 5$.

Step 4: Combine the results from steps 1-3 to find the derivative of $y = \sec^{101}(5x)$.

Using the chain rule, we have $y' = 101(\sec(5x))^{100} \cdot \sec(5x) \tan(5x) \cdot 5$.

Simplifying, we get $y' = 505(\sec(5x))^{100} \tan(5x)$.

(b) Derivative of $h(x) = \frac{10 \cos(5x)}{17}$

To find the derivative of $h(x) = \frac{10 \cos(5x)}{17}$, we will use the quotient rule. The quotient rule states that if we have a function of the form $h(x) = \frac{f(x)}{g(x)}$, then the derivative of $h$ with respect to $x$ is given by $h' = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$.

In this case, we have $h(x) = \frac{10 \cos(5x)}{17}$, which can be written as $h(x) = \frac{f(x)}{g(x)}$, where $f(x) = 10 \cos(5x)$ and $g(x) = 17$.

Step 1: Find the derivative of the numerator, which is $f(x) = 10 \cos(5x)$.

Using the chain rule, we have $f'(x) = -10 \sin(5x) \cdot 5$.

Step 2: Find the derivative of the denominator, which is $g(x) = 17$.

Using the power rule, we have $g'(x) = 0$.

Step 3: Combine the results from steps 1-2 to find the derivative of $h(x) = \frac{10 \cos(5x)}{17}$.

Using the quotient rule, we have $h' = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$.

Simplifying, we get $h' = \frac{-50 \sin(5x) \cdot 17 - 10 \cos(5x) \cdot 0}{17^2}$.

Simplifying further, we get $h' = \frac{-850 \sin(5x)}{289}$.

(c) Derivative of $y = \ln\left(12x^3 - 5x + 7\right)^{21}$

To find the derivative of $y = \ln\left(12x^3 - 5x + 7\right)^{21}$, we will use the chain rule and the power rule.

Step 1: Find the derivative of the outer function, which is $\ln(u)$.

Using the chain rule, we have $\frac{d}{dx}(\ln(u)) = \frac{1}{u} \cdot \frac{du}{dx}$.

Step 2: Find the derivative of the inner function, which is $u = (12x^3 - 5x + 7)^{21}$.

Using the power rule, we have $\frac{du}{dx} = 21(12x^3 - 5x + 7)^{20} \cdot \frac{d}{dx}(12x^3 - 5x + 7)$.

Step 3: Find the derivative of the innermost function, which is $12x^3 - 5x + 7$.

Using the power rule, we have $\frac{d}{dx}(12x^3 - 5x + 7) = 36x^2 - 5$.

Step 4: Combine the results from steps 1-3 to find the derivative of $y = \ln\left(12x^3 - 5x + 7\right)^{21}$.

Using the chain rule, we have $y' = \frac{1}{u} \cdot \frac{du}{dx}$.

Simplifying, we get $y' = \frac{21(12x^3 - 5x + 7)^{20} \cdot (36x^2 - 5)}{(12x^3 - 5x + 7)^{21}}$.

Simplifying further, we get $y' = \frac{21(36x^2 - 5)}{(12x^3 - 5x + 7)}$.

(d) Derivative of $g(x) = 200x^8 \cdot \sin(3x)$

To find the derivative of $g(x) = 200x^8 \cdot \sin(3x)$, we will use the product rule. The product rule states that if we have a function of the form $g(x) = f(x) \cdot h(x)$, then the derivative of $g$ with respect to $x$ is given by $g' = f'(x)h(x) + f(x)h'(x)$.

In this case, we have $g(x) = 200x^8 \cdot \sin(3x)$, which can be written as $g(x) = f(x) \cdot h(x)$, where $f(x) = 200x^8$ and $h(x) = \sin(3x)$.

Step 1: Find the derivative of the first function, which is $f(x) = 200x^8$.

Using the power rule, we have $f'(x) = 1600x^7$.

Step 2: Find the derivative of the second function, which is $h(x) = \sin(3x)$.

Using the chain rule, we have $h'(x) = \cos(3x) \cdot \frac{d}{dx}(3x)$.

Step 3: Find the derivative of the innermost function, which is $3x$.

Using the power rule, we have $\frac{d}{dx}(3x) = 3$.

Step 4: Combine the results from steps 1-3 to find the derivative of $g(x) = 200x^8 \cdot \sin(3x)$.

Using the product rule, we have $g' = f'(x)h(x) + f(x)h'(x)$.

Simplifying, we get $g' = 1600x^7 \sin(3x) + 200x^8 \cos(3x) \cdot 3$.

Simplifying further, we get $g' = 1600x^7 \sin(3x) + 600x^8 \cos(3x)$.

In this article, we will answer some common questions about derivatives, including what they are, how to find them, and how to use them to solve problems.

Q: What is a derivative?

A: A derivative is a measure of how a function changes as its input changes. It is a fundamental concept in calculus and is used to describe the rate of change of a function.

Q: How do I find the derivative of a function?

A: There are several rules for finding the derivative of a function, including the power rule, product rule, quotient rule, and chain rule. The power rule states that if we have a function of the form $f(x) = x^n$, then the derivative of $f$ is given by $f'(x) = nx^{n-1}$. The product rule states that if we have a function of the form $f(x) = g(x)h(x)$, then the derivative of $f$ is given by $f'(x) = g'(x)h(x) + g(x)h'(x)$. The quotient rule states that if we have a function of the form $f(x) = \frac{g(x)}{h(x)}$, then the derivative of $f$ is given by $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$. The chain rule states that if we have a function of the form $f(x) = g(h(x))$, then the derivative of $f$ is given by $f'(x) = g'(h(x)) \cdot h'(x)$.

Q: What is the difference between the derivative and the slope of a tangent line?

A: The derivative of a function at a point is the slope of the tangent line to the function at that point. However, the derivative is a measure of the rate of change of the function, while the slope of the tangent line is a measure of the rate of change of the function at a specific point.

Q: How do I use derivatives to solve problems?

A: Derivatives can be used to solve a variety of problems, including optimization problems, physics problems, and economics problems. For example, if we want to find the maximum or minimum value of a function, we can use the derivative to find the critical points of the function. If we want to model the motion of an object, we can use the derivative to find the velocity and acceleration of the object.

Q: What are some common applications of derivatives?

A: Derivatives have many applications in science, engineering, and economics. Some common applications include:

• Physics: Derivatives are used to model the motion of objects, including the velocity and acceleration of objects.
• Engineering: Derivatives are used to design and optimize systems, including electrical circuits and mechanical systems.
• Economics: Derivatives are used to model the behavior of economic systems, including the supply and demand of goods and services.
• Computer Science: Derivatives are used in machine learning and artificial intelligence to optimize algorithms and models.

Q: What are some common mistakes to avoid when working with derivatives?

A: Some common mistakes to avoid when working with derivatives include:

• Forgetting to apply the chain rule: The chain rule is a fundamental rule in calculus that is used to find the derivative of composite functions.
• Forgetting to apply the product rule: The product rule is a fundamental rule in calculus that is used to find the derivative of products of functions.
• Forgetting to apply the quotient rule: The quotient rule is a fundamental rule in calculus that is used to find the derivative of quotients of functions.
• Not checking for critical points: Critical points are points where the derivative of a function is zero or undefined. These points are important in optimization problems.

Q: What are some common derivatives that I should know?

A: Some common derivatives that you should know include:

• Derivative of x: The derivative of $x$ is $1$.
• Derivative of x^n: The derivative of $x^n$ is $nx^{n-1}$.
• Derivative of sin(x): The derivative of $\sin(x)$ is $\cos(x)$.
• Derivative of cos(x): The derivative of $\cos(x)$ is $-\sin(x)$.
• Derivative of e^x: The derivative of $e^x$ is $e^x$.

In conclusion, derivatives are a fundamental concept in calculus that are used to describe the rate of change of a function. By understanding the rules for finding derivatives and how to apply them to solve problems, you can use derivatives to model a wide range of phenomena in science, engineering, and economics.