Find The Derivative Of The Function \[$ Y \$\] Defined Implicitly In Terms Of \[$ X \$\].$\[ Y = 2 \cos(x+y) \\]$\[\frac{d Y}{d X} = \square \\](Note: Fill In The Box With The Derivative Expression.)

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Implicit Differentiation: Finding the Derivative of a Function Defined Implicitly

Implicit differentiation is a powerful technique used to find the derivative of a function defined implicitly in terms of another variable. In this article, we will explore how to find the derivative of the function y=2cos⁡(x+y)y = 2 \cos(x+y), which is defined implicitly in terms of xx. We will use the chain rule and the product rule to find the derivative of this function.

Understanding Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function that is defined implicitly in terms of another variable. This means that the function is not defined explicitly in terms of the variable, but rather in terms of another variable that is related to the first variable. In other words, the function is defined in a way that involves both the variable and another variable that is related to it.

The Chain Rule and the Product Rule

To find the derivative of the function y=2cos⁡(x+y)y = 2 \cos(x+y), we will use the chain rule and the product rule. The chain rule states that if we have a composite function of the form f(g(x))f(g(x)), then the derivative of this function is given by f′(g(x))⋅g′(x)f'(g(x)) \cdot g'(x). The product rule states that if we have a product of two functions of the form f(x)⋅g(x)f(x) \cdot g(x), then the derivative of this function is given by f′(x)⋅g(x)+f(x)⋅g′(x)f'(x) \cdot g(x) + f(x) \cdot g'(x).

Finding the Derivative of the Function

To find the derivative of the function y=2cos⁡(x+y)y = 2 \cos(x+y), we will start by differentiating the outer function, which is the cosine function. We will then use the chain rule to find the derivative of the inner function, which is the sum of xx and yy.

Step 1: Differentiate the Outer Function

The outer function is the cosine function, which has a derivative of −sin⁡(x)-\sin(x). Therefore, the derivative of the outer function is −sin⁡(x+y)-\sin(x+y).

Step 2: Use the Chain Rule to Find the Derivative of the Inner Function

The inner function is the sum of xx and yy, which can be written as x+yx+y. To find the derivative of this function, we will use the chain rule. The derivative of the inner function is given by ddx(x+y)=1+dydx\frac{d}{dx}(x+y) = 1 + \frac{dy}{dx}.

Step 3: Combine the Derivatives

Now that we have found the derivatives of the outer and inner functions, we can combine them to find the derivative of the function y=2cos⁡(x+y)y = 2 \cos(x+y). We will use the product rule to combine the derivatives.

Derivative of the Function

Using the product rule, we can write the derivative of the function y=2cos⁡(x+y)y = 2 \cos(x+y) as:

dydx=−2sin⁡(x+y)⋅(1+dydx)\frac{dy}{dx} = -2\sin(x+y) \cdot (1 + \frac{dy}{dx})

Simplifying the Derivative

To simplify the derivative, we can expand the product and combine like terms:

dydx=−2sin⁡(x+y)−2sin⁡(x+y)⋅dydx\frac{dy}{dx} = -2\sin(x+y) - 2\sin(x+y) \cdot \frac{dy}{dx}

Isolating the Derivative

To isolate the derivative, we can move all the terms involving the derivative to one side of the equation:

dydx+2sin⁡(x+y)⋅dydx=−2sin⁡(x+y)\frac{dy}{dx} + 2\sin(x+y) \cdot \frac{dy}{dx} = -2\sin(x+y)

Factoring Out the Derivative

We can factor out the derivative from the left-hand side of the equation:

dydx(1+2sin⁡(x+y))=−2sin⁡(x+y)\frac{dy}{dx}(1 + 2\sin(x+y)) = -2\sin(x+y)

Solving for the Derivative

Finally, we can solve for the derivative by dividing both sides of the equation by the factor:

dydx=−2sin⁡(x+y)1+2sin⁡(x+y)\frac{dy}{dx} = \frac{-2\sin(x+y)}{1 + 2\sin(x+y)}

In this article, we have used the chain rule and the product rule to find the derivative of the function y=2cos⁡(x+y)y = 2 \cos(x+y), which is defined implicitly in terms of xx. We have shown that the derivative of this function is given by dydx=−2sin⁡(x+y)1+2sin⁡(x+y)\frac{dy}{dx} = \frac{-2\sin(x+y)}{1 + 2\sin(x+y)}. This result demonstrates the power of implicit differentiation in finding the derivative of a function that is defined implicitly in terms of another variable.

  1. Find the derivative of the function y=3sin⁡(x+y)y = 3 \sin(x+y).
  2. Find the derivative of the function y=4cos⁡(x−y)y = 4 \cos(x-y).
  3. Find the derivative of the function y=2tan⁡(x+y)y = 2 \tan(x+y).
  1. dydx=−6cos⁡(x+y)1−6cos⁡(x+y)\frac{dy}{dx} = \frac{-6\cos(x+y)}{1 - 6\cos(x+y)}
  2. dydx=−4sin⁡(x−y)1+4sin⁡(x−y)\frac{dy}{dx} = \frac{-4\sin(x-y)}{1 + 4\sin(x-y)}
  3. dydx=2sec⁡2(x+y)1+2sec⁡2(x+y)\frac{dy}{dx} = \frac{2\sec^2(x+y)}{1 + 2\sec^2(x+y)}
  1. Stewart, J. (2016). Calculus: Early Transcendentals. Cengage Learning.
  2. Anton, H. (2018). Calculus: Early Transcendentals. John Wiley & Sons.
  3. Rogawski, J. (2019). Calculus: Early Transcendentals. W.H. Freeman and Company.
    Implicit Differentiation: Q&A

Implicit differentiation is a powerful technique used to find the derivative of a function defined implicitly in terms of another variable. In this article, we will answer some common questions about implicit differentiation and provide examples to help illustrate the concept.

Q: What is implicit differentiation?

A: Implicit differentiation is a technique used to find the derivative of a function that is defined implicitly in terms of another variable. This means that the function is not defined explicitly in terms of the variable, but rather in terms of another variable that is related to the first variable.

Q: How do I know if a function is defined implicitly?

A: A function is defined implicitly if it is not defined explicitly in terms of the variable. For example, the function y=2cos⁡(x+y)y = 2 \cos(x+y) is defined implicitly because it is not defined explicitly in terms of xx.

Q: What is the chain rule in implicit differentiation?

A: The chain rule in implicit differentiation states that if we have a composite function of the form f(g(x))f(g(x)), then the derivative of this function is given by f′(g(x))⋅g′(x)f'(g(x)) \cdot g'(x).

Q: What is the product rule in implicit differentiation?

A: The product rule in implicit differentiation states that if we have a product of two functions of the form f(x)⋅g(x)f(x) \cdot g(x), then the derivative of this function is given by f′(x)⋅g(x)+f(x)⋅g′(x)f'(x) \cdot g(x) + f(x) \cdot g'(x).

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

A: To find the derivative of a function defined implicitly, we will use the chain rule and the product rule. We will start by differentiating the outer function, which is the function that is defined explicitly in terms of the variable. We will then use the chain rule to find the derivative of the inner function, which is the function that is defined implicitly in terms of the variable.

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

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

  • Not using the chain rule and the product rule correctly
  • Not differentiating the outer function correctly
  • Not using the correct notation for the derivative
  • Not checking the domain of the function

Q: How do I check the domain of a function defined implicitly?

A: To check the domain of a function defined implicitly, we will need to find the values of the variable that make the function undefined. We will do this by setting the denominator of the function equal to zero and solving for the variable.

Q: What are some examples of functions defined implicitly?

A: Some examples of functions defined implicitly include:

  • y=2cos⁡(x+y)y = 2 \cos(x+y)
  • y=3sin⁡(x+y)y = 3 \sin(x+y)
  • y=4cos⁡(x−y)y = 4 \cos(x-y)
  • y=2tan⁡(x+y)y = 2 \tan(x+y)

Q: How do I find the derivative of a function defined implicitly using a calculator?

A: To find the derivative of a function defined implicitly using a calculator, we will need to use the implicit differentiation feature of the calculator. This feature will allow us to enter the function and the variable, and then find the derivative of the function.

Implicit differentiation is a powerful technique used to find the derivative of a function defined implicitly in terms of another variable. By understanding the chain rule and the product rule, and by following the steps outlined in this article, we can find the derivative of a function defined implicitly. We hope that this article has been helpful in answering your questions about implicit differentiation.

  1. Find the derivative of the function y=2cos⁡(x+y)y = 2 \cos(x+y).
  2. Find the derivative of the function y=3sin⁡(x+y)y = 3 \sin(x+y).
  3. Find the derivative of the function y=4cos⁡(x−y)y = 4 \cos(x-y).
  4. Find the derivative of the function y=2tan⁡(x+y)y = 2 \tan(x+y).
  1. dydx=−2sin⁡(x+y)1+2sin⁡(x+y)\frac{dy}{dx} = \frac{-2\sin(x+y)}{1 + 2\sin(x+y)}
  2. dydx=−6cos⁡(x+y)1−6cos⁡(x+y)\frac{dy}{dx} = \frac{-6\cos(x+y)}{1 - 6\cos(x+y)}
  3. dydx=−4sin⁡(x−y)1+4sin⁡(x−y)\frac{dy}{dx} = \frac{-4\sin(x-y)}{1 + 4\sin(x-y)}
  4. dydx=2sec⁡2(x+y)1+2sec⁡2(x+y)\frac{dy}{dx} = \frac{2\sec^2(x+y)}{1 + 2\sec^2(x+y)}
  1. Stewart, J. (2016). Calculus: Early Transcendentals. Cengage Learning.
  2. Anton, H. (2018). Calculus: Early Transcendentals. John Wiley & Sons.
  3. Rogawski, J. (2019). Calculus: Early Transcendentals. W.H. Freeman and Company.