9. Suppose $y=\sqrt{2x+1}$, Where $x$ And $y$ Are Functions Of $t$.a. If $\frac{dx}{dt}=3$, Find $\frac{dy}{dt}$ When $x=4$.b. If $\frac{dy}{dt}=5$, Find $\frac{dx}{dt}$

Introduction

Implicit differentiation is a powerful technique used in calculus to find the derivative of an implicitly defined function. In this article, we will explore how to use implicit differentiation to solve problems involving functions of functions. We will consider two scenarios: finding the derivative of the outer function with respect to the inner function, and finding the derivative of the inner function with respect to the outer function.

Scenario a: Finding the Derivative of the Outer Function

Problem Statement

Given the function $y=\sqrt{2x+1}$, where $x$ and $y$ are functions of $t$, find $\frac{dy}{dt}$ when $x=4$, given that $\frac{dx}{dt}=3$.

Solution

To solve this problem, we will use implicit differentiation to find the derivative of the outer function with respect to the inner function. We start by differentiating both sides of the equation $y=\sqrt{2x+1}$ with respect to $t$:

$$\frac{dy}{dt}=\frac{d}{dt}\left(\sqrt{2x+1}\right)$$

Using the chain rule, we can rewrite the right-hand side as:

$$\frac{dy}{dt}=\frac{1}{2\sqrt{2x+1}}\cdot\frac{d}{dt}\left(2x+1\right)$$

Now, we can use the fact that $\frac{dx}{dt}=3$ to substitute for $\frac{d}{dt}\left(2x+1\right)$:

$$\frac{dy}{dt}=\frac{1}{2\sqrt{2x+1}}\cdot\left(2\frac{dx}{dt}\right)$$

Substituting $x=4$ and $\frac{dx}{dt}=3$, we get:

$$\frac{dy}{dt}=\frac{1}{2\sqrt{2(4)+1}}\cdot\left(2(3)\right)=\frac{6}{2\sqrt{9}}=\frac{3}{3}=1$$

Conclusion

In this scenario, we used implicit differentiation to find the derivative of the outer function with respect to the inner function. We started by differentiating both sides of the equation $y=\sqrt{2x+1}$ with respect to $t$, and then used the chain rule and the fact that $\frac{dx}{dt}=3$ to substitute for $\frac{d}{dt}\left(2x+1\right)$. Finally, we substituted $x=4$ and $\frac{dx}{dt}=3$ to find the value of $\frac{dy}{dt}$.

Scenario b: Finding the Derivative of the Inner Function

Problem Statement

Given the function $y=\sqrt{2x+1}$, where $x$ and $y$ are functions of $t$, find $\frac{dx}{dt}$ when $x=4$, given that $\frac{dy}{dt}=5$.

Solution

To solve this problem, we will use implicit differentiation to find the derivative of the inner function with respect to the outer function. We start by differentiating both sides of the equation $y=\sqrt{2x+1}$ with respect to $t$:

$$\frac{dy}{dt}=\frac{d}{dt}\left(\sqrt{2x+1}\right)$$

Using the chain rule, we can rewrite the right-hand side as:

$$\frac{dy}{dt}=\frac{1}{2\sqrt{2x+1}}\cdot\frac{d}{dt}\left(2x+1\right)$$

Now, we can use the fact that $\frac{dy}{dt}=5$ to substitute for $\frac{dy}{dt}$:

$$5=\frac{1}{2\sqrt{2x+1}}\cdot\frac{d}{dt}\left(2x+1\right)$$

Multiplying both sides by $2\sqrt{2x+1}$, we get:

$$10\sqrt{2x+1}=\frac{d}{dt}\left(2x+1\right)$$

Using the fact that $\frac{d}{dt}\left(2x+1\right)=2\frac{dx}{dt}$, we can rewrite the right-hand side as:

$$10\sqrt{2x+1}=2\frac{dx}{dt}$$

Substituting $x=4$, we get:

$$10\sqrt{2(4)+1}=2\frac{dx}{dt}$$

Simplifying, we get:

$$10\sqrt{9}=2\frac{dx}{dt}$$

$$30=2\frac{dx}{dt}$$

Dividing both sides by 2, we get:

$$15=\frac{dx}{dt}$$

Conclusion

In this scenario, we used implicit differentiation to find the derivative of the inner function with respect to the outer function. We started by differentiating both sides of the equation $y=\sqrt{2x+1}$ with respect to $t$, and then used the chain rule and the fact that $\frac{dy}{dt}=5$ to substitute for $\frac{dy}{dt}$. Finally, we substituted $x=4$ to find the value of $\frac{dx}{dt}$.

Conclusion

$$$$$$$$$$$$$$$$$$$$

Introduction

Implicit differentiation is a powerful technique used in calculus to find the derivative of an implicitly defined function. In this article, we will explore some common questions and answers related to implicit differentiation.

Q: What is implicit differentiation?

A: Implicit differentiation is a technique used to find the derivative of an implicitly defined function. It involves differentiating both sides of an equation with respect to a variable, while treating the other variables as constants.

Q: How do I use implicit differentiation to find the derivative of an implicitly defined function?

A: To use implicit differentiation, you need to follow these steps:

1. Differentiate both sides of the equation with respect to the variable.
2. Use the chain rule to differentiate the terms that involve the other variables.
3. Simplify the resulting expression to find the derivative of the implicitly defined function.

Q: What is the chain rule in implicit differentiation?

A: The chain rule is a fundamental concept in calculus that is used to differentiate composite functions. In implicit differentiation, the chain rule is used to differentiate the terms that involve the other variables. The chain rule states that if you have a composite function of the form f(g(x)), then the derivative of the composite function is given by f'(g(x)) * g'(x).

Q: How do I use implicit differentiation to find the derivative of a function that involves a square root?

A: To use implicit differentiation to find the derivative of a function that involves a square root, you need to follow these steps:

1. Differentiate both sides of the equation with respect to the variable.
2. Use the chain rule to differentiate the terms that involve the square root.
3. Simplify the resulting expression to find the derivative of the function.

For example, if you have the function y = √(2x + 1), you can use implicit differentiation to find the derivative of the function as follows:

dy/dx = 1/(2√(2x + 1)) * d(2x + 1)/dx

Using the chain rule, you can simplify the expression to get:

dy/dx = 1/(2√(2x + 1)) * (2)

Simplifying further, you get:

dy/dx = 1/√(2x + 1)

Q: How do I use implicit differentiation to find the derivative of a function that involves a logarithm?

A: To use implicit differentiation to find the derivative of a function that involves a logarithm, you need to follow these steps:

1. Differentiate both sides of the equation with respect to the variable.
2. Use the chain rule to differentiate the terms that involve the logarithm.
3. Simplify the resulting expression to find the derivative of the function.

For example, if you have the function y = log(2x + 1), you can use implicit differentiation to find the derivative of the function as follows:

dy/dx = 1/(2x + 1) * d(2x + 1)/dx

Using the chain rule, you can simplify the expression to get:

dy/dx = 1/(2x + 1) * (2)

Simplifying further, you get:

dy/dx = 2/(2x + 1)

Q: What are some common mistakes to avoid when using implicit differentiation?

A: Some common mistakes to avoid when using implicit differentiation include:

• Forgetting to use the chain rule when differentiating terms that involve other variables.
• Not simplifying the resulting expression to find the derivative of the function.
• Making errors when differentiating the terms that involve the other variables.

Conclusion

Implicit differentiation is a powerful technique used in calculus to find the derivative of an implicitly defined function. By following the steps outlined in this article, you can use implicit differentiation to find the derivative of a wide range of functions, including those that involve square roots and logarithms. Remember to use the chain rule and simplify the resulting expression to find the derivative of the function.