Use Implicit Differentiation To Find \[$\frac{dr}{d\theta}\$\].$\[ \cos \left(r \theta^3\right) = \frac{1}{5} \\]$\[ \frac{dr}{d\theta} = \square \\]

Introduction

Implicit differentiation is a technique used to find the derivative of an implicitly defined function. In this article, we will use implicit differentiation to find the derivative of the function $\frac{dr}{d\theta}$, given the equation $\cos \left(r \theta^3\right) = \frac{1}{5}$. This technique is essential in mathematics, particularly in calculus, and has numerous applications in various fields.

What is Implicit Differentiation?

Implicit differentiation is a method used to find the derivative of an implicitly defined function. An implicitly defined function is a function that is defined by an equation, rather than an explicit formula. In other words, the function is defined implicitly, rather than explicitly. Implicit differentiation involves differentiating both sides of the equation with respect to the independent variable, while treating the dependent variable as a function of the independent variable.

The Equation

The equation we will be working with is $\cos \left(r \theta^3\right) = \frac{1}{5}$. This equation defines the relationship between the variables $r$ and $\theta$. Our goal is to find the derivative of $r$ with respect to $\theta$, denoted as $\frac{dr}{d\theta}$.

Step 1: Differentiate Both Sides of the Equation

To find the derivative of $r$ with respect to $\theta$, we will differentiate both sides of the equation with respect to $\theta$. We will use the chain rule to differentiate the left-hand side of the equation.

$$\frac{d}{d\theta} \cos \left(r \theta^3\right) = \frac{d}{d\theta} \frac{1}{5}$$

Using the chain rule, we can rewrite the left-hand side of the equation as:

$$-\sin \left(r \theta^3\right) \frac{d}{d\theta} \left(r \theta^3\right) = 0$$

Step 2: Simplify the Equation

We can simplify the equation by using the product rule to differentiate the term $r \theta^3$. The product rule states that if we have a product of two functions, $u(x)v(x)$, then the derivative of the product is given by:

$$\frac{d}{dx} \left(u(x)v(x)\right) = u'(x)v(x) + u(x)v'(x)$$

In this case, we have:

$$\frac{d}{d\theta} \left(r \theta^3\right) = r \frac{d}{d\theta} \left(\theta^3\right) + \theta^3 \frac{d}{d\theta} \left(r\right)$$

Using the power rule, we can rewrite the first term as:

$$r \frac{d}{d\theta} \left(\theta^3\right) = r \cdot 3\theta^2$$

Substituting this back into the equation, we get:

$$-\sin \left(r \theta^3\right) \left(r \cdot 3\theta^2 + \theta^3 \frac{dr}{d\theta}\right) = 0$$

**Step 3: Solve for $\frac{dr}{d\theta}$$

We can now solve for $\frac{dr}{d\theta}$ by isolating it on one side of the equation. First, we can factor out the term $\theta^3$ from the second term:

$$-\sin \left(r \theta^3\right) \left(r \cdot 3\theta^2 + \theta^3 \frac{dr}{d\theta}\right) = 0$$

$$-\sin \left(r \theta^3\right) \theta^3 \left(3r + \frac{dr}{d\theta}\right) = 0$$

Since $\theta^3$ is not equal to zero, we can divide both sides of the equation by $\theta^3$:

$$-\sin \left(r \theta^3\right) \left(3r + \frac{dr}{d\theta}\right) = 0$$

Now, we can factor out the term $3r$ from the second term:

$$-\sin \left(r \theta^3\right) \left(3r + \frac{dr}{d\theta}\right) = 0$$

$$-\sin \left(r \theta^3\right) \left(3r + \frac{dr}{d\theta}\right) = 0$$

Since $3r$ is not equal to zero, we can divide both sides of the equation by $3r$:

$$-\frac{\sin \left(r \theta^3\right)}{3r} \left(1 + \frac{dr}{d\theta}\right) = 0$$

Now, we can solve for $\frac{dr}{d\theta}$ by isolating it on one side of the equation:

$$\frac{dr}{d\theta} = -\frac{\sin \left(r \theta^3\right)}{3r}$$

Conclusion

In this article, we used implicit differentiation to find the derivative of the function $\frac{dr}{d\theta}$, given the equation $\cos \left(r \theta^3\right) = \frac{1}{5}$. We differentiated both sides of the equation with respect to $\theta$, used the chain rule and the product rule to simplify the equation, and finally solved for $\frac{dr}{d\theta}$. The result is a derivative that is expressed in terms of the variables $r$ and $\theta$. This technique is essential in mathematics, particularly in calculus, and has numerous applications in various fields.

References

• [1] Thomas, G. B. (2010). Calculus and Analytic Geometry. Addison-Wesley.
• [2] Stewart, J. (2012). Calculus: Early Transcendentals. Cengage Learning.
• [3] Anton, H. (2012). Calculus: A New Horizon. John Wiley & Sons.

Further Reading

• Implicit differentiation is a powerful tool for finding derivatives of implicitly defined functions.
• It is essential to understand the chain rule and the product rule in order to use implicit differentiation effectively.
• Implicit differentiation has numerous applications in various fields, including physics, engineering, and economics.
  Implicit Differentiation: A Q&A Guide =====================================

Introduction

Implicit differentiation is a powerful technique used to find the derivative of an implicitly defined function. In our previous article, we used implicit differentiation to find the derivative of the function $\frac{dr}{d\theta}$, given the equation $\cos \left(r \theta^3\right) = \frac{1}{5}$. In this article, we will answer some of the most frequently asked questions about implicit differentiation.

Q: What is implicit differentiation?

A: Implicit differentiation is a technique used to find the derivative of an implicitly defined function. An implicitly defined function is a function that is defined by an equation, rather than an explicit formula.

Q: When should I use implicit differentiation?

A: You should use implicit differentiation when you are given an equation that defines a relationship between two or more variables, and you need to find the derivative of one of the variables with respect to another variable.

Q: How do I use implicit differentiation?

A: To use implicit differentiation, you need to differentiate both sides of the equation with respect to the independent variable, while treating the dependent variable as a function of the independent variable. You will need to use the chain rule and the product rule to simplify the equation.

Q: What is the chain rule?

A: The chain rule is a rule used to differentiate composite functions. It states that if we have a composite function of the form $f(g(x))$, then the derivative of the composite function is given by:

$$\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$$

Q: What is the product rule?

A: The product rule is a rule used to differentiate products of functions. It states that if we have a product of two functions, $u(x)v(x)$, then the derivative of the product is given by:

$$\frac{d}{dx} \left(u(x)v(x)\right) = u'(x)v(x) + u(x)v'(x)$$

Q: How do I apply the chain rule and the product rule in implicit differentiation?

A: To apply the chain rule and the product rule in implicit differentiation, you need to differentiate both sides of the equation with respect to the independent variable, while treating the dependent variable as a function of the independent variable. You will need to use the chain rule and the product rule to simplify the equation.

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

A: Some common mistakes to avoid in implicit differentiation include:

• Not differentiating both sides of the equation with respect to the independent variable
• Not treating the dependent variable as a function of the independent variable
• Not using the chain rule and the product rule to simplify the equation
• Not isolating the derivative of the dependent variable on one side of the equation

Q: How do I check my work in implicit differentiation?

A: To check your work in implicit differentiation, you need to:

• Verify that you have differentiated both sides of the equation with respect to the independent variable
• Verify that you have treated the dependent variable as a function of the independent variable
• Verify that you have used the chain rule and the product rule to simplify the equation
• Verify that you have isolated the derivative of the dependent variable on one side of the equation

Conclusion

Implicit differentiation is a powerful technique used to find the derivative of an implicitly defined function. In this article, we have answered some of the most frequently asked questions about implicit differentiation. We hope that this article has been helpful in clarifying the concepts and techniques involved in implicit differentiation.

References

• [1] Thomas, G. B. (2010). Calculus and Analytic Geometry. Addison-Wesley.
• [2] Stewart, J. (2012). Calculus: Early Transcendentals. Cengage Learning.
• [3] Anton, H. (2012). Calculus: A New Horizon. John Wiley & Sons.

Further Reading

• Implicit differentiation is a powerful tool for finding derivatives of implicitly defined functions.
• It is essential to understand the chain rule and the product rule in order to use implicit differentiation effectively.
• Implicit differentiation has numerous applications in various fields, including physics, engineering, and economics.