Let $h(t) = \tan(2x + 6$\].Then $h'(1$\] Is $\square$ And $h''(1$\] Is $\square$.

Let $h(t) = \tan(2t + 6)$: Finding the First and Second Derivatives

In this article, we will explore the concept of finding the first and second derivatives of a given function. We will use the function $h(t) = \tan(2t + 6)$ as an example to demonstrate the process. The first derivative of a function represents the rate of change of the function with respect to the variable, while the second derivative represents the rate of change of the first derivative.

To find the first derivative of the function $h(t) = \tan(2t + 6)$, we will use the chain rule of differentiation. The chain rule states that if we have a composite function of the form $f(g(x))$, then the derivative of the function is given by $f'(g(x)) \cdot g'(x)$.

In this case, we have $h(t) = \tan(2t + 6)$, which can be written as $f(g(t))$, where $f(u) = \tan(u)$ and $g(t) = 2t + 6$. Using the chain rule, we can find the first derivative of $h(t)$ as follows:

$$h'(t) = f'(g(t)) \cdot g'(t)$$

$$h'(t) = \sec^2(2t + 6) \cdot 2$$

$$h'(t) = 2\sec^2(2t + 6)$$

Now, we need to find the value of $h'(1)$.

Finding $h'(1)$

To find the value of $h'(1)$, we need to substitute $t = 1$ into the expression for $h'(t)$.

$$h'(1) = 2\sec^2(2(1) + 6)$$

$$h'(1) = 2\sec^2(8)$$

Using a calculator, we can find the value of $\sec^2(8)$ and multiply it by 2 to get the final answer.

To find the second derivative of the function $h(t) = \tan(2t + 6)$, we will use the chain rule and the product rule of differentiation. The product rule states that if we have a function of the form $f(x)g(x)$, then the derivative of the function is given by $f'(x)g(x) + f(x)g'(x)$.

In this case, we have $h'(t) = 2\sec^2(2t + 6)$, which can be written as $f(t)g(t)$, where $f(t) = 2$ and $g(t) = \sec^2(2t + 6)$. Using the product rule, we can find the second derivative of $h(t)$ as follows:

$$h''(t) = f'(t)g(t) + f(t)g'(t)$$

$$h''(t) = 0 \cdot \sec^2(2t + 6) + 2 \cdot 2\sec^2(2t + 6)\tan(2t + 6)$$

$$h''(t) = 4\sec^2(2t + 6)\tan(2t + 6)$$

Now, we need to find the value of $h''(1)$.

Finding $h''(1)$

To find the value of $h''(1)$, we need to substitute $t = 1$ into the expression for $h''(t)$.

$$h''(1) = 4\sec^2(2(1) + 6)\tan(2(1) + 6)$$

$$h''(1) = 4\sec^2(8)\tan(8)$$

Using a calculator, we can find the value of $\sec^2(8)$ and $\tan(8)$ and multiply them by 4 to get the final answer.

In this article, we have found the first and second derivatives of the function $h(t) = \tan(2t + 6)$. We have used the chain rule and the product rule of differentiation to find the derivatives. We have also found the values of $h'(1)$ and $h''(1)$ by substituting $t = 1$ into the expressions for the derivatives. The first derivative represents the rate of change of the function with respect to the variable, while the second derivative represents the rate of change of the first derivative.
Q&A: Let $h(t) = \tan(2t + 6)$: Finding the First and Second Derivatives

In our previous article, we explored the concept of finding the first and second derivatives of a given function. We used the function $h(t) = \tan(2t + 6)$ as an example to demonstrate the process. In this article, we will answer some frequently asked questions related to the topic.

Q: What is the first derivative of the function $h(t) = \tan(2t + 6)$?

A: The first derivative of the function $h(t) = \tan(2t + 6)$ is given by $h'(t) = 2\sec^2(2t + 6)$.

Q: How do I find the value of $h'(1)$?

A: To find the value of $h'(1)$, you need to substitute $t = 1$ into the expression for $h'(t)$. This gives you $h'(1) = 2\sec^2(8)$.

Q: What is the second derivative of the function $h(t) = \tan(2t + 6)$?

A: The second derivative of the function $h(t) = \tan(2t + 6)$ is given by $h''(t) = 4\sec^2(2t + 6)\tan(2t + 6)$.

Q: How do I find the value of $h''(1)$?

A: To find the value of $h''(1)$, you need to substitute $t = 1$ into the expression for $h''(t)$. This gives you $h''(1) = 4\sec^2(8)\tan(8)$.

Q: What is the significance of the first and second derivatives of a function?

A: The first derivative of a function represents the rate of change of the function with respect to the variable. The second derivative represents the rate of change of the first derivative. This information can be useful in various applications, such as physics, engineering, and economics.

Q: How do I use the chain rule and the product rule of differentiation?

A: The chain rule states that if we have a composite function of the form $f(g(x))$, then the derivative of the function is given by $f'(g(x)) \cdot g'(x)$. The product rule states that if we have a function of the form $f(x)g(x)$, then the derivative of the function is given by $f'(x)g(x) + f(x)g'(x)$. These rules can be used to find the derivatives of various functions.

Q: What are some common applications of derivatives in real-life situations?

A: Derivatives have many applications in real-life situations, such as:

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

In this article, we have answered some frequently asked questions related to the topic of finding the first and second derivatives of a given function. We have also provided examples and explanations to help illustrate the concepts. We hope that this article has been helpful in clarifying any doubts you may have had.