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Chapter 3: Problem 23

Expand or simplify to compute the following: \(\frac{d}{d x}((x+1)(x+1)(x-1)(x-1))\)

Short Answer

Expert verified
The derivative of the function \((x+1)(x+1)(x-1)(x-1)\) is \(4x^3 - 6x^2 + 4x\).

Step by step solution

01

Simplify the Given Equation

First of all, simplify the given equation. It is given that the function \(f(x) = (x+1)(x+1)(x-1)(x-1)\). This can be simplified as \(f(x) = (x^2 + 2x + 1)(x^2 - 2x + 1)\). This is the original function in its simplest form.
02

Differentiate the Simplified Function

Next, find the derivative of the function. To find the derivative of the function, use the product rule, which is \[(uv)' = u'v + uv'\] where \(u = x^2 + 2x + 1\) and \(v = x^2 - 2x + 1\). Applying the product rule, we get: \[f'(x) = (2x+2)(x^2 - 2x + 1) + (x^2 + 2x + 1)(2x - 2)\] This is the derivative of the original function.
03

Simplify the Derivative

The last step is to simplify the derivative: \[f'(x) = 2x^3 - 4x^2 + 2x + 2x^2 - 4x + 2 + 2x^3 - 4x^2 + 2x - 2\] After simplifying, we get: \[f'(x) = 4x^3 - 6x^2 + 4x\]

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Key Concepts

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

Product Rule in Calculus Differentiation
In calculus, the product rule is essential when differentiating the product of two or more functions. If you have functions multiplied together, like in the exercise, you can't simply take the derivative of each individually and multiply them. Instead, the product rule comes to our rescue.

Imagine two functions, denoted by \( u(x) \) and \( v(x) \). To find the derivative of their product, you apply:
  • Differentiate \( u(x) \) to get \( u'(x) \).
  • Differentiate \( v(x) \) to get \( v'(x) \).
  • Then, the product rule formula is: \((uv)' = u'v + uv'\).
This rule uses both derivatives of these functions and ensures we correctly account for how both functions contribute to the rate of change of the product. It's important to remember each term must be differentiated with this rule when you encounter such problems in calculus.
Polynomial Differentiation
Differentiating polynomials is a fundamental part of calculus. A polynomial is made up of terms like \( ax^n \), where \( a \) is a coefficient, \( x \) is a variable, and \( n \) is a non-negative integer exponent. The differentiation process involves applying a power rule.

The power rule states that when you differentiate \( x^n \), you multiply by the exponent and reduce the exponent by one. For example, the derivative of \( x^2 \) is \( 2x \). Each term in a polynomial gets this treatment:
  • If you have \( 7x^5 \), its derivative is \( 35x^4 \).
  • For a constant term like 9, its derivative is 0 since constants don鈥檛 change.
  • Linear terms like \( 3x \) become just the coefficient, 3.
By simplifying each term this way, the overall derivative quickly takes shape. It鈥檚 a straightforward yet powerful process that prepares you for more complex differentiation tasks.
Simplifying Expressions After Differentiation
Once you've applied calculus rules like the product rule to find a derivative, the next step is often to simplify the result. Simplification involves removing unnecessary terms and combining like terms to achieve a more manageable expression.

Usually, derivatives can appear intimidating with many terms multiplied or added. Here's how you simplify:
  • Carefully distribute any products across sums or differences.
  • Combine terms with the same degree, which means adding up the coefficients of similar powers of \( x \).
  • Simplify constants whenever possible for ease of expression.
This final step ensures that the derivative is in its simplest, most understandable form. Simplification doesn't change the function's meaning but does make it easier to analyze, graph, or use in further computations.

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