Problem 21 - Thomas Calculus

Question

Find the derivatives of the functions in Exercises \(v=(1-t)\left(1+t^{2}\right)^{-1}\)

Step-by-Step Solution

Answer

The derivative is \( v'(t) = \frac{t^2 - 2t - 1}{(1+t^2)^2} \).

Step 1: Identify the Structure of the Function

The function given is \( v = (1-t)(1+t^2)^{-1} \). This is a product of two functions: \( u(t) = 1-t \) and \( w(t) = (1+t^2)^{-1} \). We will differentiate these using the product rule.

Step 2: Differentiate the First Function

Differentiate \( u(t) = 1 - t \). The derivative of \( u(t) \), which is \( u'(t) \), is \( -1 \) because the derivative of a constant is zero and the derivative of \( t \) is 1.

Step 3: Differentiate the Second Function

Differentiate \( w(t) = (1 + t^2)^{-1} \) using the chain rule. Consider it as \((1 + t^2)^{-1}\). The derivative of \((1+t^2)^{-1}\) is \(- (1 + t^2)^{-2} \times 2t = -2t(1 + t^2)^{-2}\).

Step 4: Apply the Product Rule

The product rule states: \( (uv)' = u'v + uv' \). Substitute \( u(t) = 1-t \), \( w(t) = (1+t^2)^{-1} \), \( u'(t) = -1 \), and \( w'(t) = -2t(1+t^2)^{-2} \).

Step 5: Calculate the Derivative

Using the product rule, we get \( v'(t) = (-1)(1+t^2)^{-1} + (1-t)(-2t)(1+t^2)^{-2} \). Simplify this expression to get \( v'(t) = -\frac{1}{1+t^2} - \frac{2t(1-t)}{(1+t^2)^2} \).

Step 6: Simplify the Expression

Combine the terms over a common denominator to obtain \( v'(t) = \frac{- (1 + t^2) - 2t(1-t)}{(1+t^2)^2} \). Simplify the numerator to yield \( v'(t) = \frac{-1 -t^2 -2t + 2t^2}{(1+t^2)^2} = \frac{t^2 - 2t - 1}{(1+t^2)^2} \).

Key Concepts

product rule

The product rule is a fundamental concept in calculus used for differentiating functions that are multiplied together. If you have two functions, say \( f(x) \) and \( g(x) \), their derivative is not as simple as just multiplying the derivatives of each function. The product rule provides a formula for correctly finding the derivative of the product of two functions.

It is formulated as:

• \((fg)' = f'g + fg'\)

Here, \(f'\) and \(g'\) are the derivatives of \(f\) and \(g\), respectively. The rule says you take the derivative of the first function, multiply it by the second, and then add it to the first function multiplied by the derivative of the second.

This is crucial for equations like \(v=(1-t)(1+t^2)^{-1} \), where both components need to be differentiated correctly using this rule. Remember that each part requires detailed attention to ensure accuracy in calculations.

chain rule

The chain rule is another essential tool in calculus, especially when dealing with composite functions, where you have a function inside another function. It's like peeling an onion, where you take one layer off at a time. To apply the chain rule, you identify the outer function and the inner function and differentiate each part systematically.

For a function represented as \(y = f(g(x))\), the chain rule states:

• \(\frac{dy}{dx} = f'(g(x)) \, g'(x)\)

This means you differentiate the outer function at the inner function (with respect to the outer variable), and then multiply by the derivative of the inner function.

Applying this to our example, when differentiating \((1+t^2)^{-1}\), we first notice the outer function is \(x^{-1}\) and the inner function is \(1+t^2\). We differentiate \(x^{-1}\) to get \(-x^{-2}\), apply it to \(1+t^2\), and then multiply by the derivative of \(1+t^2\), which is \(2t\). Thus the derivative becomes \(-2t(1+t^2)^{-2}\). Always be cautious of the layers of each function and differentiate methodically.

function differentiation

Function differentiation is the process of finding the derivative of a function, which tells us the rate at which the function's value changes with respect to a change in its input value. This is a core principle in calculus and is applied extensively to solve a variety of mathematical problems.

In the context of the given function \(v = (1-t)(1+t^2)^{-1}\), we utilize both the product and chain rules to find the derivative.

Here's the step-by-step for differentiating our function:

• Separate the function into parts you can individually differentiate, i.e., \(u(t) = 1-t\) and \(w(t) = (1+t^2)^{-1}\).
• Differentiate each part using appropriate rules: \(u'(t) = -1\) and \(w'(t) = -2t(1+t^2)^{-2}\).
• Apply the product rule: \(v'(t) = u'(t)w(t) + u(t)w'(t)\).
• Simplify the resulting expression to ensure clarity and precision.

While differentiating, pay attention to constants, power rules, and the implications of the product and chain rules to expertly handle the calculations. This approach ensures you can tackle complex functions by breaking them down into manageable parts.