Question 5. Find F'(x)Solution. (a) F(x) = In Arc Tan (2x) (b) F(x) = F(x)= Ex Sechx
---
Introduction
In calculus, differentiation is a fundamental concept that allows us to analyze the rate of change of functions. Determining the derivatives of various functions is crucial in fields such as physics, engineering, economics, and many more. This article focuses on solving two specific differentiation problems involving composite and hyperbolic functions.
The problems are as follows:
- Part (a): Find the derivative of \( F(x) = \ln(\arctan(2x)) \).
- Part (b): Find the derivative of \( F(x) = e^x \operatorname{sech} x \).
Understanding how to differentiate these functions requires a good grasp of chain rule, product rule, and properties of hyperbolic functions. We will delve into each problem step-by-step, providing detailed explanations and relevant formulas to enhance comprehension.
---
Part (a): Differentiating \( F(x) = \ln(\arctan(2x)) \)
Understanding the function
The function \( F(x) = \ln(\arctan(2x)) \) involves a composition of functions:
- The natural logarithm function \( \ln u \),
- The inverse tangent (arctangent) function \( \arctan v \),
- The linear function \( 2x \).
Hence, differentiating this function involves applying the chain rule multiple times.
Step-by-step differentiation
- Recognize the composite structure
Let:
- \( u = \arctan(2x) \),
- So, \( F(x) = \ln u \).
The derivative of \( \ln u \) with respect to \( u \) is \( \frac{1}{u} \).
- Differentiate \( u = \arctan(2x) \)
Using the chain rule:
- \( \frac{d}{dx} \arctan(2x) = \frac{1}{1 + (2x)^2} \times \frac{d}{dx}(2x) = \frac{1}{1 + 4x^2} \times 2 = \frac{2}{1 + 4x^2} \).
- Combine the derivatives
Applying the chain rule:
\[
F'(x) = \frac{1}{u} \times \frac{du}{dx} = \frac{1}{\arctan(2x)} \times \frac{2}{1 + 4x^2}.
\]
Final derivative expression:
\[
\boxed{
F'(x) = \frac{2}{(1 + 4x^2) \cdot \arctan(2x)}.
}
\]
Additional notes
- The derivative is valid for all \( x \) where \( \arctan(2x) \neq 0 \) (i.e., \( x \neq 0 \)), as the logarithm is undefined for zero or negative arguments.
- Understanding the domain is crucial when applying derivatives involving logarithmic functions.
---
Part (b): Differentiating \( F(x) = e^x \operatorname{sech} x \)
Understanding the function
The function \( F(x) = e^x \operatorname{sech} x \) is a product of exponential and hyperbolic secant functions. To differentiate it, we need to apply the product rule, which states:
\[
\frac{d}{dx}[u(x) \cdot v(x)] = u'(x) v(x) + u(x) v'(x).
\]
Step-by-step differentiation
- Identify \( u(x) \) and \( v(x) \)
Let:
- \( u(x) = e^x \),
- \( v(x) = \operatorname{sech} x \).
- Differentiate each component
- \( u'(x) = e^x \),
- \( v(x) = \operatorname{sech} x \).
Recall that:
\[
\operatorname{sech} x = \frac{1}{\cosh x}.
\]
The derivative of \( \operatorname{sech} x \) is:
\[
\frac{d}{dx} \operatorname{sech} x = - \operatorname{sech} x \tanh x,
\]
which can be derived from the quotient rule or known hyperbolic derivatives.
- Apply the product rule
\[
F'(x) = u'(x) v(x) + u(x) v'(x) = e^x \operatorname{sech} x + e^x \times \left( - \operatorname{sech} x \tanh x \right).
\]
Simplify:
\[
F'(x) = e^x \operatorname{sech} x - e^x \operatorname{sech} x \tanh x.
\]
Factor out common terms:
\[
F'(x) = e^x \operatorname{sech} x (1 - \tanh x).
\]
Final derivative expression:
\[
\boxed{
F'(x) = e^x \operatorname{sech} x (1 - \tanh x).
}
\]
Additional notes
- The hyperbolic tangent \( \tanh x \) is related to \( \sinh x \) and \( \cosh x \) as:
\[
\tanh x = \frac{\sinh x}{\cosh x}.
\]
- The hyperbolic secant \( \operatorname{sech} x \) is always positive and decreases as \( |x| \) increases.
- The expression \( 1 - \tanh x \) influences the sign and magnitude of the derivative across different \( x \) values.
---
Summary and Key Takeaways
Differentiation Techniques Used
- Chain Rule: Essential for differentiating compositions like \( \ln(\arctan(2x)) \).
- Product Rule: Necessary for differentiating products such as \( e^x \operatorname{sech} x \).
- Hyperbolic Function Derivatives: Understanding derivatives of hyperbolic functions like \( \operatorname{sech} x \) and \( \tanh x \) is crucial.
Final Results Recap
- Part (a):
\[
\boxed{
F'(x) = \frac{2}{(1 + 4x^2) \cdot \arctan(2x)}.
}
\]
- Part (b):
\[
\boxed{
F'(x) = e^x \operatorname{sech} x (1 - \tanh x).
}
\]
Applications of These Derivatives
Understanding these derivatives has broad applications:
- In physics, for calculating rates of change involving inverse trigonometric functions and hyperbolic functions.
- In engineering, for analyzing signal processing involving hyperbolic functions.
- In mathematics, for solving differential equations and modeling phenomena with these functions.
---
Conclusion
Differentiating complex functions like \( \ln(\arctan(2x)) \) and \( e^x \operatorname{sech} x \) showcases the power of calculus tools such as the chain rule, product rule, and properties of hyperbolic functions. Mastering these techniques enables one to analyze a wide range of mathematical models and real-world problems involving advanced functions.
By carefully applying the rules and understanding the properties of inverse and hyperbolic functions, students and professionals can confidently tackle similar differentiation problems, enhancing their analytical skills and mathematical proficiency.
