Problem 44 - Single Variable Calculus

Question

find the derivative of the function. \(g(x)=\tanh ^{-1} \frac{x}{2}\)

Step-by-Step Solution

Answer

The derivative of the function \(g(x) = \tanh^{-1} \frac{x}{2}\) is \(g'(x) = \frac{1}{2(1 - (\frac{x}{2})^2)}\).

Step 1: Recognize the Inverse Hyperbolic Tangent Function

In this step, we notice that g(x) is given as the arctanh of a function. So we have: \(g(x) = arctanh(\frac{x}{2})\) Step 2: Apply implicit differentiation

Step 2: Implicit Differentiation

To differentiate the function implicitly, we first express the function in terms of y and x using the definition of the inverse hyperbolic tangent: \(\tanh(g(x)) = \frac{x}{2}\) Step 3: Differentiate with respect to x

Step 3: Differentiate with Respect to x

Now, we will differentiate both sides of the equation with respect to x. On the left side, we will use the chain rule and on the right side, we will differentiate with respect to x: \(\frac{d}{dx}[\tanh(g(x))] = \frac{d}{dx}[\frac{x}{2}]\) Step 4: Apply chain rule and simplify

Step 4: Chain Rule and Simplification

Applying the chain rule on the left side and differentiating the right side with respect to x, we get: \(\frac{d\tanh(g(x))}{dx} \cdot \frac{dg(x)}{dx} = \frac{1}{2}\) Using the fact that the derivative of \(\tanh(u)\) is \((1 - \tanh^2(u)) \cdot \frac{du}{dx}\), we get: \((1 - \tanh^2(g(x))) \cdot \frac{dg(x)}{dx} = \frac{1}{2}\) Recall that \(\tanh(g(x)) = \frac{x}{2}\), so we can substitute back: \((1 - (\frac{x}{2})^2) \cdot \frac{dg(x)}{dx} = \frac{1}{2}\) Step 5: Solve for the derivative

Step 5: Solve for the Derivative

We now need to solve the equation for \(\frac{dg(x)}{dx}\): \(\frac{dg(x)}{dx} = \frac{\frac{1}{2}}{1 - (\frac{x}{2})^2}\) Finally, \(\frac{dg(x)}{dx} = \frac{1}{2(1 - (\frac{x}{2})^2)}\) So, the derivative of \(g(x) = arctanh(\frac{x}{2})\) is \(\frac{dg(x)}{dx} = \frac{1}{2(1 - (\frac{x}{2})^2)}\).

Key Concepts

Implicit Differentiation

Implicit differentiation is a powerful technique often used when dealing with functions that are not explicitly defined in terms of one variable, such as inverse functions.
For instance, when you have an equation involving two variables, like in the case of hyperbolic functions, you may not be able to solve for one variable in terms of the other easily. Implicit differentiation allows you to find the derivative of one variable with respect to the other by differentiating both sides of the equation with respect to the independent variable and then isolating the derivative of the dependent variable.

In the exercise, to find the derivative of the inverse hyperbolic tangent function, we first wrote the function implicitly as \(\tanh(g(x)) = \frac{x}{2}\). The next step was to differentiate both sides with respect to \(x\) while treating \(g(x)\) as a function of \(x\). This is where the chain rule comes into play, which brings us to the next pivotal concept.

Chain Rule

The chain rule is a fundamental theorem in calculus, used when differentiating composite functions. Whenever you encounter a function nested within another function, like \(\tanh(g(x))\), the chain rule guides you in teasing apart the derivatives of the inner and outer layers.

Mathematically, if \(f\) is a function and \(g\) is another function, then the derivative of their composite \(f(g(x))\) is given by \(f'(g(x))\cdot g'(x)\). In the exercise, we had to apply this rule to differentiate \(\tanh(g(x))\) with respect to \(x\), which involves first taking the derivative of the outer function \(\tanh(u)\) and then multiplying it by the derivative \(g'(x)\) of the inner function \(g(x)\).

Recognizing when to use the chain rule is crucial in tackling derivatives of complex function compositions.

Hyperbolic Functions Calculus

Hyperbolic functions are analogues to the trigonometric functions but for hyperbolas instead of circles. They appear frequently in various branches of mathematics, including calculus.

Some of the basic hyperbolic functions include \(\sinh(x)\), representing hyperbolic sine, and \(\cosh(x)\), representing hyperbolic cosine. Just like their trigonometric counterparts, they have inverse functions—for example, \(\tanh^{-1}(x)\) is the inverse hyperbolic tangent function.

These functions have properties and derivatives that are often analogous to trigonometric functions, with some key differences. For example, the derivative of \(\tanh(x)\) is \(1 - \tanh^2(x)\), which we used in the step-by-step solution to find \(\frac{dg(x)}{dx}\). Understanding these properties is crucial when working with hyperbolic functions in calculus.

Derivative of Inverse Trigonometric Functions

Derivatives of inverse trigonometric functions can seem daunting, but once you grasp the connection with their basic trigonometric forms, it becomes much easier.

The derivatives of these functions typically involve the original function itself and the use of the square root in the expressions. For example, the derivative of \(\arcsin(x)\) is \(\frac{1}{\sqrt{1-x^2}}\), and for \(\arctan(x)\), it is \(\frac{1}{1+x^2}\). These derivatives are key when working with related rates and integrals that involve arc trigonometric functions.

In our exercise, although we dealt with a hyperbolic function, the concept of finding the derivative of its inverse form is parallel to that of inverse trigonometric functions—both involve implicit differentiation and often require the chain rule for their solution.