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

Short Answer

Expert verified
\( \frac{dy}{dt} = -\text{sech}^2 \left( \frac{1}{t} \right) + 2t \tanh \left( \frac{1}{t} \right) \).

Step by step solution

01

Apply Product Rule

The function is of the form \( y = u(t) \cdot v(t) \), where \( u(t) = t^2 \) and \( v(t) = \tanh \left( \frac{1}{t} \right) \). By the product rule, the derivative \( \frac{dy}{dt} \) is given by \( u(t) \cdot \frac{d}{dt}[v(t)] + v(t) \cdot \frac{d}{dt}[u(t)] \).
02

Differentiate \(u(t) = t^2\)

The derivative of \( u(t) = t^2 \) with respect to \( t \) is \( 2t \).
03

Differentiate \(v(t) = \tanh \left( \frac{1}{t} \right)\)

Utilize the chain rule to differentiate \( v(t) = \tanh \left( \frac{1}{t} \right) \). The outer function is \( \tanh(x) \) which has a derivative \( \text{sech}^2(x) \), and the inner function is \( \frac{1}{t} \) which has a derivative of \( -\frac{1}{t^2} \). Thus, \( \frac{d}{dt} \left[ \tanh \left( \frac{1}{t} \right) \right] = \text{sech}^2 \left( \frac{1}{t} \right) \cdot \left(-\frac{1}{t^2}\right) \).
04

Apply the Results to Product Rule

Substitute the derivatives back into the product rule formula: \[ \frac{dy}{dt} = t^2 \cdot \left( \text{sech}^2 \left( \frac{1}{t} \right) \cdot \left(-\frac{1}{t^2}\right) \right) + \tanh \left( \frac{1}{t} \right) \cdot (2t) \].
05

Simplify the Expression

Simplify the expression from Step 4: \[ \frac{dy}{dt} = t^2 \cdot \left( -\frac{1}{t^2} \text{sech}^2 \left( \frac{1}{t} \right) \right) + 2t \cdot \tanh \left( \frac{1}{t} \right) \]. This reduces to: \[ -\text{sech}^2 \left( \frac{1}{t} \right) + 2t \tanh \left( \frac{1}{t} \right) \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule
When finding the derivative of functions that are multiplied together, the product rule is your go-to tool. Essentially, if you have two functions, say \( u(t) \) and \( v(t) \), both dependent on \( t \), the product rule helps you find their combined derivative. Here's how it works in a simple way:

- **Formula**: To get the derivative of their product, which is \( u(t) \cdot v(t) \), you calculate it as follows: \[ \frac{d}{dt}[u(t) \cdot v(t)] = u(t) \cdot \frac{d}{dt}[v(t)] + v(t) \cdot \frac{d}{dt}[u(t)] \]
- **Practically**: You first take one function, say \( u(t) \), multiply it by the derivative of the other function \( v(t) \), and then add the product of the other way around: \( v(t) \) times the derivative of \( u(t) \).

In our example, when dealing with \( y = t^2 \cdot \tanh \frac{1}{t} \), the product rule is essential. You differentiate each part according to the rule and combine them. This step-by-step process helps in managing complex functions effortlessly by breaking them into easier pieces.
Chain Rule
The chain rule is crucial when you need to differentiate composite functions, where one function sits inside another. Think of it as a way to "peel layers."

- **How It Works**: Imagine \( y = f(g(t)) \), where \( g(t) \) is within \( f \). The chain rule states that the derivative \( \frac{dy}{dt} \) is given by: \[ \frac{dy}{dt} = f'(g(t)) \cdot g'(t) \]
Here, \( f'(g(t)) \) denotes the derivative of the outer function evaluated at the inner function, and \( g'(t) \) is the derivative of the inner function itself.

- **Example in Practice**: For \( \tanh \left( \frac{1}{t} \right) \), the hyperbolic tangent is the outer layer, while the reciprocal (\( \frac{1}{t} \)) is the inner layer. You first differentiate the hyperbolic tangent to get \( \text{sech}^2 \), and then multiply it by the derivative of \( \frac{1}{t} \), which is \( -\frac{1}{t^2} \).
The chain rule allows complex nesting of functions to be easily handled by separately dealing with their layers.
Hyperbolic Functions
Hyperbolic functions, like \( \text{tanh} \), \( \text{sech} \), \( \text{sinh} \), and \( \text{cosh} \), are analogs to the classical trigonometric functions, but they arise from hyperbolas instead of circles.

- **Understanding \( \tanh(x) \)**: The hyperbolic tangent function, \( \tanh(x) \), is similar to the regular tangent but describes the ratio of hyperbolic sine and hyperbolic cosine: \[ \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \]
- **Derivatives**: Important for calculus, the derivative of \( \tanh(x) \) is \( \text{sech}^2(x) \), where \( \text{sech}(x) \) is the hyperbolic secant. - For differentiating composite forms like \( \tanh \left( \frac{1}{t} \right) \), you apply the chain rule.

These functions often model real-world phenomena such as electrical currents and population dynamics because of their smooth, incremental nature.

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