Problem 16 - Thomas Calculus

Question

In Exercises \(13-24,\) find the derivative of \(y\) with respect to the appropriate variable. $$y=t^{2} \tanh \frac{1}{t}$$

Step-by-Step Solution

Answer

The derivative is \( y'(t) = 2t \tanh\left(\frac{1}{t}\right) - \text{sech}^2\left(\frac{1}{t}\right) \).

Step 1: Identify the Structure

The function given is a product of two functions: \( y = t^2 \cdot \tanh\left(\frac{1}{t}\right) \). This suggests that the product rule will be useful for differentiation.

Step 2: Recall the Product Rule

The product rule states that if you have two functions \( u(t) \) and \( v(t) \), then the derivative \( (uv)' \) is given by \( u'v + uv' \). Here, let \( u(t) = t^2 \) and \( v(t) = \tanh\left(\frac{1}{t}\right) \).

Step 3: Differentiate the First Function \(u(t)\)

Differentiate \( u(t) = t^2 \) with respect to \( t \). The derivative is \( u'(t) = 2t \).

Step 4: Differentiate the Second Function \(v(t)\)

Differentiate \( v(t) = \tanh\left(\frac{1}{t}\right) \) using the chain rule. Let \( w = \frac{1}{t} \), so \( v(t) = \tanh(w) \). The derivative of \( \tanh(w) \) with respect to \( w \) is \( \text{sech}^2(w) \), and \( \frac{dw}{dt} = -\frac{1}{t^2} \). Thus, \( v'(t) = \text{sech}^2\left(\frac{1}{t}\right) \cdot \left(-\frac{1}{t^2}\right) = -\frac{\text{sech}^2\left(\frac{1}{t}\right)}{t^2} \).

Step 5: Apply the Product Rule

Using the product rule, we get\[ y'(t) = 2t \cdot \tanh\left(\frac{1}{t}\right) + t^2 \cdot \left(-\frac{\text{sech}^2\left(\frac{1}{t}\right)}{t^2}\right) \].

Step 6: Simplify the Expression

Simplify the expression obtained by applying the product rule:\[ y'(t) = 2t \tanh\left(\frac{1}{t}\right) - \text{sech}^2\left(\frac{1}{t}\right) \]

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus differentiation. It's applied when you need to find the derivative of a product of two functions. Suppose you have two functions, \( u(t) \) and \( v(t) \), then the derivative of their product is given by the formula:\[ (uv)' = u'v + uv' \]This rule helps you differentiate expressions like \( y = t^2 \cdot \tanh\left(\frac{1}{t}\right) \). Here, you set \( u(t) = t^2 \) and \( v(t) = \tanh\left(\frac{1}{t}\right) \).

• First, differentiate \( u(t) \) with respect to \( t \), resulting in \( u'(t) = 2t \).
• Then, differentiate \( v(t) \) using another rule.

This sets you up to apply the product rule effectively. Calculate the two derivatives separately so you can plug them into the formula to find the overall derivative of the product.

Chain Rule

The chain rule is crucial when differentiating composite functions. Consider a function \( v(t) = \tanh\left(\frac{1}{t}\right) \). It's composite because it's a hyperbolic tangent involving another function, \( w = \frac{1}{t} \). The chain rule states:\[ \frac{dv}{dt} = \frac{dv}{dw} \cdot \frac{dw}{dt} \]Here’s how you apply the chain rule:

• Differentiate \( \tanh(w) \) with respect to \( w \), giving \( \text{sech}^2(w) \).
• Then, find \( \frac{dw}{dt} \) where \( w = \frac{1}{t} \), which is \( -\frac{1}{t^2} \).

Combine these to get the derivative \( v'(t) = \text{sech}^2\left(\frac{1}{t}\right) \cdot -\frac{1}{t^2} \). Understanding chain rule will help you with any function that’s a bit complex internally.

Hyperbolic Functions

Hyperbolic functions, like \( \tanh(x) \), are analogous to trigonometric functions but for hyperbolas as opposed to circles. The function \( \tanh(x) \) is defined as:\[ \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \]where \( \sinh(x) = \frac{e^x - e^{-x}}{2} \) and \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).

• \( \tanh(x) \) is essential in various calculus problems, especially those involving derivatives.
• Its derivative, \( \text{sech}^2(x) \), appears often in differentiation tasks, including in the current problem with \( \tanh\left(\frac{1}{t}\right) \).

Grasping hyperbolic functions will help smoothly transition through both calculus and physics problems where these functions frequently arise.

Derivatives

Derivatives represent the rate of change of a function. If you have a function \( y(t) = t^2 \tanh\left(\frac{1}{t}\right) \), finding \( y'(t) \) gives you how \( y \) changes concerning \( t \).

• The derivative of a constant times a function, such as \( t^2 \), is straightforward once you know basic rules like the product and the chain rule.
• Combining different rules of derivatives, like in the step-by-step solution, can simplify seemingly complicated problems.

This process not only involves mathematical formulas but also enhances the understanding of how variables in calculus behave dynamically. Knowing how to find derivatives is essential for any calculus work, as it forms the foundation for more complex studies like integration and differential equations.