Problem 51 - Calculus Single Variable

Question

Differentiate the given expression with respect to \(x\). $$ \tanh ^{-1}(\cos (x)) $$

Step-by-Step Solution

Answer

The derivative is \(-\csc(x)\).

Step 1: Identify the Function to Differentiate

The given expression is the inverse hyperbolic tangent of cosine: \( \tanh^{-1} (\cos(x)) \). We need to differentiate this function with respect to \( x \).

Step 2: Differentiate the Outer Function Using the Chain Rule

Start by differentiating the outer function, \( \tanh^{-1}(u) \), where \( u = \cos(x) \). The derivative of \( \tanh^{-1}(u) \) with respect to \( u \) is \( \frac{1}{1-u^2} \). Thus, we have \( \frac{d}{dx}[ \tanh^{-1}(\cos(x)) ] = \frac{1}{1-(\cos(x))^2} \cdot \frac{du}{dx} \).

Step 3: Differentiate the Inner Function

Now differentiate \( \cos(x) \) with respect to \( x \), which gives \( \frac{du}{dx} = -\sin(x) \).

Step 4: Apply the Chain Rule

Combine the results using the chain rule: \( \frac{d}{dx}[ \tanh^{-1}(\cos(x)) ] = \frac{1}{1-(\cos(x))^2} \cdot (-\sin(x)) \).

Step 5: Simplify the Expression Further

Since \( 1 - (\cos(x))^2 = (\sin(x))^2 \) (from the Pythagorean identity), we get \( \frac{-\sin(x)}{(\sin(x))^2} = \frac{-1}{\sin(x)} \).

Step 6: Final Simplification

Thus, the expression simplifies to \( -\csc(x) \), where \( \csc(x) = \frac{1}{\sin(x)} \).

Key Concepts

Chain Rule

Understanding the chain rule is essential when dealing with compositions of functions. It allows us to find the derivative of composite functions efficiently. The chain rule states that if you have a function composed of two functions, say \( f(g(x)) \), the derivative is: \[ f'(g(x)) \cdot g'(x) \] This means you first find the derivative of the outer function \( f \), evaluated at the inner function \( g(x) \), and then multiply it by the derivative of the inner function \( g \). In the provided exercise, we applied the chain rule to \( \tanh^{-1}(\cos(x)) \). Here, \( \tanh^{-1}(u) \) is the outer function and \( \cos(x) \) is the inner function. By differentiating each part separately and combining them, we successfully find the derivative. Remember, the chain rule is powerful for handling nested functions, and mastering it opens up ways to differentiate more complex functions easily.

Inverse Hyperbolic Functions

Inverse hyperbolic functions, like \( \tanh^{-1}(u) \), are essential in various branches of mathematics, including calculus and complex analysis. These functions are similar to their trigonometric counterparts but defined for hyperbolic functions. For instance, the inverse hyperbolic tangent, \( \tanh^{-1}(u) \), is defined such that: \[ y = \tanh^{-1}(u) \quad \Rightarrow \quad u = \tanh(y) \] Its derivative with respect to \( u \) is: \[ \frac{d}{du} \tanh^{-1}(u) = \frac{1}{1-u^2} \] This formula mirrors the derivative of \( \tan^{-1}(u) \) in its structure but applies to the hyperbolic domain. In our exercise, we recognized \( \tanh^{-1}(u) \) as the outer function, differentiated it initially concerning \( u \), and applied the chain rule to complete the expression. Inverse hyperbolic functions are crucial for solving integrals and differential equations that involve hyperbolic terms.

Trigonometric Functions

Trigonometric functions like \( \cos(x) \) and \( \sin(x) \) are base blocks in calculus, appearing frequently in various applications from physics to engineering. When differentiating trigonometric functions, it is key to remember the exponential derivatives: - The derivative of \( \cos(x) \) is \( -\sin(x) \).- The identity \( 1 - \cos^2(x) = \sin^2(x) \) is used often, such as in our exercise.These identities and formulas are intuitive tools aimed at simplification in calculus. In the exercise, knowing that the derivative of \( \cos(x) \) is \( -\sin(x) \) helped simplify the final expression to \( -\csc(x) \), where \( \csc(x) = \frac{1}{\sin(x)} \). Having a firm grasp on differentiating trigonometric functions and using their identities allows for navigating through more complicated calculus problems with ease.