Problem 28 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\left(\theta^{2}+2 \theta\right) \tanh ^{-1}(\theta+1)$$

Step-by-Step Solution

Answer

The derivative is \((2\theta + 2)\tanh^{-1}(\theta+1) - 1\).

Step 1: Identify the Rule to Apply

The function given is a product of two functions: \(u(\theta) = \theta^2 + 2\theta\) and \(v(\theta) = \tanh^{-1}(\theta+1)\). To find the derivative, we will use the Product Rule, which states that \((uv)' = u'v + uv'\).

Step 2: Differentiate \(u(\theta)\)

Differentiate \(u(\theta) = \theta^2 + 2\theta\):\[ u'(\theta) = \frac{d}{d\theta}(\theta^2 + 2\theta) = 2\theta + 2 \].

Step 3: Differentiate \(v(\theta)\)

Differentiate \(v(\theta) = \tanh^{-1}(\theta+1)\). Recall that the derivative of \(\tanh^{-1}(x)\) is \(\frac{1}{1-x^2}\), but we also apply the chain rule:\[ v'(\theta) = \frac{d}{d\theta}\tanh^{-1}(\theta+1) = \frac{1}{1-(\theta+1)^2} \cdot 1 = \frac{1}{1-(\theta^2+2\theta+1)} = \frac{1}{-\theta(\theta+2)} \].

Step 4: Apply the Product Rule

Using the product rule formula \((uv)' = u'v + uv'\), substitute the derivatives:\[ \frac{dy}{d\theta} = (2\theta + 2)\tanh^{-1}(\theta+1) + (\theta^2 + 2\theta)\cdot\frac{1}{-\theta(\theta+2)} \].

Step 5: Simplify the Derivative

Simplify the expression:1. \( (2\theta + 2)\tanh^{-1}(\theta+1) \) stays as it is.2. Simplifying \( (\theta^2 + 2\theta)\cdot\frac{1}{-\theta(\theta+2)} \) involves combining terms: \( \frac{\theta^2 + 2\theta}{-\theta(\theta+2)} = -(1) \). Therefore, the simplified derivative is:\[ \frac{dy}{d\theta} = (2\theta + 2)\tanh^{-1}(\theta+1) - 1 \].

Key Concepts

Product Rule

When finding the derivative of a product of two functions, the Product Rule is an essential tool. The general principle of the Product Rule is described by the formula:

• If you have a function that is the product of two other functions, say \( u(x) \) and \( v(x) \), then the derivative of their product \( uv \) with respect to \( x \) is given by \( (uv)' = u'v + uv' \).

This means we differentiate \( u(x) \) to find \( u'(x) \), differentiate \( v(x) \) to find \( v'(x) \), and then apply the formula.

In our exercise, you apply the Product Rule to two functions: \( u(\theta) = \theta^2 + 2\theta \) and \( v(\theta) = \tanh^{-1}(\theta+1) \). It’s essential to first identify and differentiate each of these functions separately. Once we have both derivatives, we substitute them back into the Product Rule formula to find the overall derivative.

Chain Rule

The Chain Rule is another crucial rule for differentiation, especially when dealing with composite functions.

• A composite function is a function within another function, such as \( f(g(x)) \).
• The Chain Rule helps us find the derivative of a composite function by stating that: \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \).

The Chain Rule consists of taking the derivative of the outer function, evaluated at the inner function, and then multiplying by the derivative of the inner function.

In the context of our exercise, we're handling the inverse hyperbolic function \( \tanh^{-1}(\theta + 1) \). Here, \( f(x) = \tanh^{-1}(x) \) and \( g(\theta) = \theta + 1 \), thus creating a composite function. This requires us to use the Chain Rule to find \( v'(\theta) \), the derivative of \( v(\theta) = \tanh^{-1}(\theta+1) \). When we apply the Chain Rule, we recognize that \( \frac{d}{d\theta} \tanh^{-1}(\theta + 1) = \frac{1}{1-(\theta+1)^2} \cdot 1 \), which simplifies the process of finding the derivative.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are used frequently in calculus and appear in many formulas like the one seen in our exercise. The most common ones are \( \sinh^{-1}, \cosh^{-1}, \text{and } \tanh^{-1} \). They are similar to inverse trigonometric functions but involve hyperbolic functions.

• The inverse hyperbolic tangent, \( \tanh^{-1}(x) \), is defined for the range \(-1 < x < 1\) and has a distinct derivative formula: \( \frac{d}{dx} \tanh^{-1}(x) = \frac{1}{1-x^2} \).

Understanding this derivative is key when working with inverse hyperbolic functions.

In the problem at hand, it’s crucial to correctly apply this formula when differentiating \( \tanh^{-1}(\theta + 1) \). Notice how the formula is modified by the Chain Rule due to the argument \( \theta + 1 \). Mastery of these functions supports deeper comprehension of calculus problems and ensures successful application of the derivative formulas correctly in various contexts.