Problem 22 - Calculus Early Transcendentals

Question

Find \(d y /\left.d x\right|_{x=1}\). \(y=\left(2 x^{7}-x^{2}\right)\left(\frac{x-1}{x+1}\right)\)

Step-by-Step Solution

Answer

The derivative \(dy/dx\) at \(x=1\) is \(\frac{1}{2}\).

Step 1: Identify the Product Rule

The given function is a product of two functions: \(u(x) = 2x^7 - x^2\) and \(v(x) = \frac{x-1}{x+1}\). We will use the product rule for differentiation, which states that \(\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\).

Step 2: Differentiate the First Function

Differentiate \(u(x) = 2x^7 - x^2\) with respect to \(x\).\[u'(x) = \frac{d}{dx}(2x^7) - \frac{d}{dx}(x^2) = 14x^6 - 2x\]

Step 3: Differentiate the Second Function

Differentiate \(v(x) = \frac{x-1}{x+1}\) using the quotient rule, which states \(\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\).Let \(f(x) = x-1\) and \(g(x) = x+1\), then:\[\begin{align*}v'(x) & = \frac{(1)(x+1) - (x-1)(1)}{(x+1)^2} \ & = \frac{x + 1 - x + 1}{(x+1)^2} \ & = \frac{2}{(x+1)^2}\end{align*}\]

Step 4: Apply the Product Rule

Substitute the derivatives into the product rule:\[\begin{align*}\frac{dy}{dx} & = u'(x)v(x) + u(x)v'(x) \ & = (14x^6 - 2x) \cdot \frac{x-1}{x+1} + (2x^7 - x^2) \cdot \frac{2}{(x+1)^2}\end{align*}\]

Step 5: Evaluate at \(x=1\)

Substitute \(x = 1\) into the expression for \(\frac{dy}{dx}\). For the first term:\[\begin{align*}\frac{dy}{dx}\bigg|_{x=1} & = (14(1)^6 - 2(1)) \cdot \frac{1-1}{1+1} + (2(1)^7 - (1)^2) \cdot \frac{2}{(1+1)^2} \ & = (14 - 2) \cdot \frac{0}{2} + (2 - 1) \cdot \frac{2}{4}\end{align*}\]For the second term:\[1 \cdot \frac{1}{2} = \frac{1}{2}\]Thus, \(\frac{dy}{dx}\bigg|_{x=1} = \frac{1}{2}\).

Step 6: Conclusion

The derivative of \(y\) with respect to \(x\) evaluated at \(x=1\) is \(\frac{1}{2}\).

Key Concepts

Product Rule

The product rule is a fundamental principle in calculus used when you want to differentiate the product of two functions. If you have a function expressed as a product of two separate functions, say \( u(x) \) and \( v(x) \), then the product rule states:

• \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \)

This formula essentially tells us that to find the derivative of a product, you need to:- Differentiate the first function \( u(x) \), denoted as \( u'(x) \).- Multiply \( u'(x) \) by the second function \( v(x) \).- Differentiate the second function \( v(x) \), denoted as \( v'(x) \).- Multiply \( u(x) \) by \( v'(x) \).- Finally, add these two results together to get the derivative of the product.
This rule is particularly useful and often essential when working with polynomials multiplied by other functions or any expression that can't be simplified into a single variable term.

Quotient Rule

The quotient rule is another rule for differentiation used when you want to find the derivative of a quotient of two functions. If you have a function that's written as \( \frac{f(x)}{g(x)} \), then the quotient rule is applied as follows:

• \( \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \)

Here's what each term represents:- \( f'(x) \) is the derivative of the numerator function \( f(x) \).- \( g'(x) \) is the derivative of the denominator function \( g(x) \).- The numerator of the derivative (\( f'(x)g(x) - f(x)g'(x) \)) combines these derivatives in a specific way: you multiply the derivative of the first function by the second function and subtract the first function multiplied by the derivative of the second function.- The denominator of the derivative is the square of \( g(x) \).
This can be quite useful especially when dealing with rational functions, helping to systematically and accurately find their derivatives.

Differentiation

Differentiation is a core concept in calculus centered around determining the rate at which a function is changing at any given point. When you differentiate a function, you're essentially finding its derivative, which tells you how the function's value changes as its input changes. Differentiation is used widely across calculus and applied mathematics for various analyses.To perform differentiation, one needs to apply rules like the product and quotient rules, among others, depending on how complex the function is:

• Power Rule: For functions of the form \( f(x) = x^n \), the derivative is \( f'(x) = nx^{n-1} \).
• Constant Rule: The derivative of a constant is always zero.
• Sum Rule: If a function is the sum of two functions, \( f(x) = u(x) + v(x) \), then the derivative is the sum of the derivatives: \( f'(x) = u'(x) + v'(x) \).

Differentiation allows you to find tangents to curves, optimize functions, and solve real-world engineering and physics problems by providing a mathematical way to describe change.