The Vector-valued Function H H H Is Defined By H ( T ) = ⟨ E T + E − T , E T − E − T ⟩ H(t)=\left\langle E^t+e^{-t}, E^t-e^{-t}\right\rangle H ( T ) = ⟨ E T + E − T , E T − E − T ⟩ . Which Of The Following Is H ′ ( Ln ⁡ 3 H^{\prime}(\ln 3 H ′ ( Ln 3 ]?A. 4 5 \frac{4}{5} 5 4 ​ B. 5 4 \frac{5}{4} 4 5 ​ C.

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Introduction

In this article, we will explore the concept of vector-valued functions and their derivatives. Specifically, we will examine the function $h(t)=\left\langle e^t+e^{-t}, e^t-e^{-t}\right\rangle$ and find its derivative at the point $t=\ln 3$. We will use the definition of a derivative and the chain rule to find the derivative of $h(t)$ and then evaluate it at the given point.

The Vector-Valued Function $h(t)$

The vector-valued function $h(t)$ is defined as $h(t)=\left\langle e^t+e^{-t}, e^t-e^{-t}\right\rangle$. This function takes a real number $t$ as input and returns a vector in the plane. The first component of the vector is $e^t+e^{-t}$, and the second component is $e^t-e^{-t}$.

Finding the Derivative of $h(t)$

To find the derivative of $h(t)$, we will use the definition of a derivative. The derivative of a function $f(t)$ is defined as $f^{\prime}(t)=\lim_{h\to 0}\frac{f(t+h)-f(t)}{h}$. We will apply this definition to the function $h(t)$.

Let $h(t)=\left\langle e^t+e^{-t}, e^t-e^{-t}\right\rangle$. Then, we have:

$$h^{\prime}(t)=\lim_{h\to 0}\frac{h(t+h)-h(t)}{h}$$

$$=\lim_{h\to 0}\frac{\left\langle e^{t+h}+e^{-(t+h)}, e^{t+h}-e^{-(t+h)}\right\rangle-\left\langle e^t+e^{-t}, e^t-e^{-t}\right\rangle}{h}$$

$$=\lim_{h\to 0}\frac{\left\langle e^t e^h+e^{-t}e^{-h}, e^t e^h-e^{-t}e^{-h}\right\rangle-\left\langle e^t+e^{-t}, e^t-e^{-t}\right\rangle}{h}$$

$$=\lim_{h\to 0}\frac{\left\langle (e^t+e^{-t})e^h-(e^t+e^{-t})+(e^t-e^{-t})e^{-h}-(e^t-e^{-t}), (e^t+e^{-t})e^h-(e^t+e^{-t})-(e^t-e^{-t})e^{-h}+(e^t-e^{-t})\right\rangle}{h}$$

$$=\lim_{h\to 0}\frac{\left\langle (e^t+e^{-t})(e^h-1)+(e^t-e^{-t})(e^{-h}-1), (e^t+e^{-t})(e^h-1)-(e^t-e^{-t})(e^{-h}-1)\right\rangle}{h}$$

Applying the Chain Rule

We can simplify the expression for $h^{\prime}(t)$ by applying the chain rule. The chain rule states that if we have a composite function $f(g(t))$, then the derivative of $f(g(t))$ is given by $f^{\prime}(g(t))g^{\prime}(t)$. We can apply this rule to the expression for $h^{\prime}(t)$.

Let $f(x)=\left\langle x, x\right\rangle$. Then, we have:

$$f^{\prime}(x)=\left\langle 1, 1\right\rangle$$

Now, let $g(t)=e^t+e^{-t}$. Then, we have:

$$g^{\prime}(t)=e^t-e^{-t}$$

We can now apply the chain rule to the expression for $h^{\prime}(t)$:

$$h^{\prime}(t)=f^{\prime}(g(t))g^{\prime}(t)$$

$$=\left\langle 1, 1\right\rangle(e^t-e^{-t})$$

$$=\left\langle e^t-e^{-t}, e^t-e^{-t}\right\rangle$$

Evaluating $h^{\prime}(\ln 3)$

We are asked to find the value of $h^{\prime}(\ln 3)$. To do this, we will substitute $t=\ln 3$ into the expression for $h^{\prime}(t)$.

$$h^{\prime}(\ln 3)=\left\langle e^{\ln 3}-e^{-\ln 3}, e^{\ln 3}-e^{-\ln 3}\right\rangle$$

$$=\left\langle 3-1/3, 3-1/3\right\rangle$$

$$=\left\langle \frac{8}{3}, \frac{8}{3}\right\rangle$$

Conclusion

In this article, we have found the derivative of the vector-valued function $h(t)=\left\langle e^t+e^{-t}, e^t-e^{-t}\right\rangle$. We have used the definition of a derivative and the chain rule to find the derivative of $h(t)$ and then evaluated it at the point $t=\ln 3$. The value of $h^{\prime}(\ln 3)$ is $\left\langle \frac{8}{3}, \frac{8}{3}\right\rangle$.

Discussion

The vector-valued function $h(t)$ is a simple example of a function that takes a real number as input and returns a vector in the plane. The derivative of $h(t)$ is a vector-valued function that represents the rate of change of $h(t)$ with respect to $t$. In this article, we have used the definition of a derivative and the chain rule to find the derivative of $h(t)$ and then evaluated it at the point $t=\ln 3$.

Final Answer

The final answer is $\boxed{\left\langle \frac{8}{3}, \frac{8}{3}\right\rangle}$.

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Introduction

In our previous article, we explored the concept of vector-valued functions and their derivatives. Specifically, we examined the function $h(t)=\left\langle e^t+e^{-t}, e^t-e^{-t}\right\rangle$ and found its derivative at the point $t=\ln 3$. In this article, we will answer some common questions related to the vector-valued function $h(t)$ and its derivative.

Q: What is the vector-valued function $h(t)$?

A: The vector-valued function $h(t)$ is defined as $h(t)=\left\langle e^t+e^{-t}, e^t-e^{-t}\right\rangle$. This function takes a real number $t$ as input and returns a vector in the plane.

Q: What is the derivative of the vector-valued function $h(t)$?

A: The derivative of the vector-valued function $h(t)$ is given by $h^{\prime}(t)=\left\langle e^t-e^{-t}, e^t-e^{-t}\right\rangle$.

Q: How do you find the derivative of a vector-valued function?

A: To find the derivative of a vector-valued function, you can use the definition of a derivative and the chain rule. The definition of a derivative states that the derivative of a function $f(t)$ is given by $f^{\prime}(t)=\lim_{h\to 0}\frac{f(t+h)-f(t)}{h}$. The chain rule states that if you have a composite function $f(g(t))$, then the derivative of $f(g(t))$ is given by $f^{\prime}(g(t))g^{\prime}(t)$.

Q: What is the value of $h^{\prime}(\ln 3)$?

A: The value of $h^{\prime}(\ln 3)$ is $\left\langle \frac{8}{3}, \frac{8}{3}\right\rangle$.

Q: What is the significance of the vector-valued function $h(t)$?

A: The vector-valued function $h(t)$ is a simple example of a function that takes a real number as input and returns a vector in the plane. The derivative of $h(t)$ is a vector-valued function that represents the rate of change of $h(t)$ with respect to $t$.

Q: How do you use the vector-valued function $h(t)$ in real-world applications?

A: The vector-valued function $h(t)$ can be used in a variety of real-world applications, such as modeling the motion of an object in a plane, representing the velocity of an object, and calculating the acceleration of an object.

Q: What are some common mistakes to avoid when working with vector-valued functions?

A: Some common mistakes to avoid when working with vector-valued functions include:

• Not using the correct notation for vector-valued functions
• Not applying the chain rule correctly
• Not evaluating the derivative at the correct point
• Not using the correct units for the derivative

Conclusion

In this article, we have answered some common questions related to the vector-valued function $h(t)$ and its derivative. We have discussed the definition of a vector-valued function, the derivative of a vector-valued function, and the significance of the vector-valued function $h(t)$. We have also provided some tips for using the vector-valued function $h(t)$ in real-world applications and some common mistakes to avoid when working with vector-valued functions.

Final Answer

The final answer is $\boxed{\left\langle \frac{8}{3}, \frac{8}{3}\right\rangle}$.