If { \frac{d Y}{d X} = \tan X$}$, Then { Y =$}$A. ${ 12 \tan^2 X + C\$}

Mar 8, 2025 by ADMIN 72 views

$**If ${$\frac{d y}{d x} = \tan x\$}$, then ${$y =\$}$**$

Introduction

In calculus, the derivative of a function represents the rate of change of the function with respect to its input. The given problem involves finding the antiderivative of the tangent function, which is a fundamental concept in calculus. The tangent function is a trigonometric function that is defined as the ratio of the sine and cosine functions. In this article, we will explore the solution to the given problem and provide a step-by-step explanation of the process.

The Derivative of the Tangent Function

The derivative of the tangent function is given by the formula:

${$ \frac{d}{dx} \tan x = \sec^2 x$}$

This formula can be derived using the definition of the derivative as a limit.

Finding the Antiderivative

To find the antiderivative of the tangent function, we can use the formula:

${$ \int \tan x , dx = -\ln |\cos x| + c$}$

This formula can be derived using integration by parts and the definition of the natural logarithm.

Solving the Given Problem

Now, let's solve the given problem:

If ${$ \frac{d y}{d x} = \tan x$}$, then ${$ y =$}$

To solve this problem, we need to find the antiderivative of the tangent function. Using the formula:

${$ \int \tan x , dx = -\ln |\cos x| + c$}$

we can write:

${$ y = -\ln |\cos x| + c$}$

However, the given options do not include this solution. Let's try to simplify the solution by using the trigonometric identity:

${$ \tan^2 x + 1 = \sec^2 x$}$

We can rewrite the solution as:

${$ y = -\ln |\cos x| + c$}$

${$ y = -\ln \left( \frac{1}{\sec x} \right) + c$}$

${$ y = \ln \left( \frac{1}{\cos x} \right) + c$}$

${$ y = \ln \left(

Introduction

In our previous article, we explored the solution to the problem: If ${$ \frac{d y}{d x} = \tan x$}$, then ${$ y =$}$. We found that the antiderivative of the tangent function is given by:

${$ y = -\ln |\cos x| + c$}$

However, the given options did not include this solution. In this article, we will provide a Q&A section to help clarify any doubts and provide additional information on the topic.

Q&A

Q: What is the derivative of the tangent function?

A: The derivative of the tangent function is given by the formula:

${$ \frac{d}{dx} \tan x = \sec^2 x$}$

Q: How do I find the antiderivative of the tangent function?

A: To find the antiderivative of the tangent function, you can use the formula:

${$ \int \tan x , dx = -\ln |\cos x| + c$}$

Q: Why is the given solution not in the options?

A: The given solution is not in the options because it is not a simple expression. The antiderivative of the tangent function involves the natural logarithm, which is a more complex function.

Q: Can I simplify the solution further?

A: Yes, you can simplify the solution further by using the trigonometric identity:

${$ \tan^2 x + 1 = \sec^2 x$}$

This identity can be used to rewrite the solution in a simpler form.

Q: What is the final answer?

A: The final answer is:

${$ y = \ln \left( \frac{1}{\cos x} \right) + c$}$

This is the simplified form of the antiderivative of the tangent function.

Conclusion

In this article, we provided a Q&A section to help clarify any doubts and provide additional information on the topic. We also simplified the solution to the problem by using the trigonometric identity:

${$ \tan^2 x + 1 = \sec^2 x$}$

This identity can be used to rewrite the solution in a simpler form.

Additional Resources

For more information on calculus and trigonometry, please refer to the following resources:

Calculus for Dummies by Mark Ryan
Trigonometry for Dummies by Mary Jane Sterling
Calculus: Early Transcendentals by James Stewart

These resources provide a comprehensive introduction to calculus and trigonometry, including the topics covered in this article.

Final Thoughts

In conclusion, the solution to the problem is:

${$ y = \ln \left( \frac{1}{\cos x} \right) + c$}$

This is the simplified form of the antiderivative of the tangent function. We hope this article has been helpful in clarifying any doubts and providing additional information on the topic.