Understanding the Derivative of a Product Function

The question asks us to find the derivative of a function \(y\) which is a product of several terms, and then evaluate this derivative at a specific point, \(x = 0\).

The function is given by:

\(y = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)(1 + x^{16})\)

We need to find \(\frac{dy}{dx}\) and then determine the value of \(\frac{dy}{dx}\) when \(x = 0\).

There are a couple of ways to approach this problem involving the derivative of a product. We can use the product rule directly, or we can simplify the expression for \(y\) first before differentiating.

Method 1: Using the Product Rule and Evaluating at \(x=0\)

Let the function \(y\) be written as a product of five functions:

\(y = f_1(x) \cdot f_2(x) \cdot f_3(x) \cdot f_4(x) \cdot f_5(x)\)

Where:

• \(f_1(x) = 1 + x\)
• \(f_2(x) = 1 + x^2\)
• \(f_3(x) = 1 + x^4\)
• \(f_4(x) = 1 + x^8\)
• \(f_5(x) = 1 + x^{16}\)

The product rule for the derivative of a product of multiple functions states that:

\(\frac{dy}{dx} = f_1'(x)f_2f_3f_4f_5 + f_1f_2'f_3f_4f_5 + f_1f_2f_3'f_4f_5 + f_1f_2f_3f_4'f_5 + f_1f_2f_3f_4f_5'\)

Where the arguments \((x)\) are omitted for brevity, but each \(f_i\) and \(f_i'\) is evaluated at \(x\).

Finding the Derivatives of Each Factor

Now, let's find the derivative of each function \(f_i(x)\) with respect to \(x\):

• \(f_1'(x) = \frac{d}{dx}(1 + x) = 1\)
• \(f_2'(x) = \frac{d}{dx}(1 + x^2) = 2x\)
• \(f_3'(x) = \frac{d}{dx}(1 + x^4) = 4x^3\)
• \(f_4'(x) = \frac{d}{dx}(1 + x^8) = 8x^7\)
• \(f_5'(x) = \frac{d}{dx}(1 + x^{16}) = 16x^{15}\)

Evaluating Functions and Derivatives at \(x = 0\)

We need to find \(\frac{dy}{dx}\) at \(x = 0\). Let's evaluate each \(f_i(x)\) and \(f_i'(x)\) at \(x = 0\):

• \(f_1(0) = 1 + 0 = 1\)
• \(f_2(0) = 1 + 0^2 = 1\)
• \(f_3(0) = 1 + 0^4 = 1\)
• \(f_4(0) = 1 + 0^8 = 1\)
• \(f_5(0) = 1 + 0^{16} = 1\)

And the derivatives at \(x = 0\):

• \(f_1'(0) = 1\)
• \(f_2'(0) = 2(0) = 0\)
• \(f_3'(0) = 4(0)^3 = 0\)
• \(f_4'(0) = 8(0)^7 = 0\)
• \(f_5'(0) = 16(0)^{15} = 0\)

Substituting Values into the Product Rule

Now, substitute these values into the product rule formula evaluated at \(x = 0\):

\(\left.\frac{dy}{dx}\right|_{x=0} = f_1'(0)f_2(0)f_3(0)f_4(0)f_5(0) + f_1(0)f_2'(0)f_3(0)f_4(0)f_5(0) + f_1(0)f_2(0)f_3'(0)f_4(0)f_5(0) + f_1(0)f_2(0)f_3(0)f_4'(0)f_5(0) + f_1(0)f_2(0)f_3(0)f_4(0)f_5'(0)\)

Plugging in the values we calculated:

\(\left.\frac{dy}{dx}\right|_{x=0} = (1)(1)(1)(1)(1) + (1)(0)(1)(1)(1) + (1)(1)(0)(1)(1) + (1)(1)(1)(0)(1) + (1)(1)(1)(1)(0)\)

\(\left.\frac{dy}{dx}\right|_{x=0} = 1 + 0 + 0 + 0 + 0\)

\(\left.\frac{dy}{dx}\right|_{x=0} = 1\)

Using the product rule and evaluating at \(x=0\) directly gives us the result 1.

Method 2: Recognizing a Pattern and Simplifying First

The product expression resembles the expansion of a difference of squares. Consider the product \((1-x)(1+x) = 1-x^2\), \((1-x^2)(1+x^2) = 1-x^4\), and so on.

Let the given function be \(y\). Multiply \(y\) by \((1-x)\):

\(y(1-x) = (1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})\)

Using the difference of squares repeatedly:

• \((1-x)(1+x) = 1 - x^2\)
• \((1-x^2)(1+x^2) = 1 - x^4\)
• \((1-x^4)(1+x^4) = 1 - x^8\)
• \((1-x^8)(1+x^8) = 1 - x^{16}\)
• \((1-x^{16})(1+x^{16}) = 1 - x^{32}\)

So, the product simplifies to:

\(y(1-x) = 1 - x^{32}\)

Provided \(x \neq 1\), we can write \(y\) as:

\(y = \frac{1 - x^{32}}{1 - x}\)

Now, we can differentiate \(y\) with respect to \(x\) using the quotient rule, which states that if \(y = \frac{u}{v}\), then \(\frac{dy}{dx} = \frac{u'v - uv'}{v^2}\).

Let \(u = 1 - x^{32}\) and \(v = 1 - x\).

• \(u' = \frac{d}{dx}(1 - x^{32}) = -32x^{31}\)
• \(v' = \frac{d}{dx}(1 - x) = -1\)

Applying the quotient rule:

\(\frac{dy}{dx} = \frac{(-32x^{31})(1 - x) - (1 - x^{32})(-1)}{(1 - x)^2}\)

\(\frac{dy}{dx} = \frac{-32x^{31} + 32x^{32} + 1 - x^{32}}{(1 - x)^2}\)

\(\frac{dy}{dx} = \frac{1 - 32x^{31} + 31x^{32}}{(1 - x)^2}\)

Evaluating the Derivative at \(x = 0\)

Now, substitute \(x = 0\) into the expression for \(\frac{dy}{dx}\):

\(\left.\frac{dy}{dx}\right|_{x=0} = \frac{1 - 32(0)^{31} + 31(0)^{32}}{(1 - 0)^2}\)

\(\left.\frac{dy}{dx}\right|_{x=0} = \frac{1 - 0 + 0}{(1)^2}\)

\(\left.\frac{dy}{dx}\right|_{x=0} = \frac{1}{1}\)

\(\left.\frac{dy}{dx}\right|_{x=0} = 1\)

Conclusion

Both methods yield the same result. The value of \(\frac{dy}{dx}\) at \(x = 0\) is 1.

The final answer is \(\mathbf{1}\).

Revision Table: Key Calculus Concepts

Additional Information: Why Evaluating at x=0 Simplified Method 1

In Method 1, when we evaluated the derivative expression at \(x = 0\), notice that most terms became zero. This happened because the derivatives of \(1+x^2, 1+x^4, 1+x^8,\) and \(1+x^{16}\) all contain \(x\) raised to a positive power (\(2x, 4x^3, 8x^7, 16x^{15}\)). When \(x=0\), any term multiplied by \(x\) to a positive power becomes zero.

The product rule result is:

\(\frac{dy}{dx} = (1)(1+x^2)(1+x^4)(1+x^8)(1+x^{16}) + (1+x)(2x)(1+x^4)(1+x^8)(1+x^{16}) + \cdots\)

At \(x=0\):

\(\left.\frac{dy}{dx}\right|_{x=0} = (1)(1+0)(1+0)(1+0)(1+0) + (1+0)(2 \cdot 0)(1+0)(1+0)(1+0) + \cdots\)

\(\left.\frac{dy}{dx}\right|_{x=0} = (1)(1)(1)(1)(1) + (1)(0)(1)(1)(1) + (1)(1)(0)(1)(1) + (1)(1)(1)(0)(1) + (1)(1)(1)(1)(0)\)

\(\left.\frac{dy}{dx}\right|_{x=0} = 1 + 0 + 0 + 0 + 0 = 1\)

Only the term where the derivative of \((1+x)\) appears (which is 1, a constant) and the other factors are evaluated at \(x=0\) (which are all 1) contributes to the final sum at \(x=0\). This demonstrates how plugging in the specific value *before* fully simplifying can sometimes make calculations easier.