Polynomial expantion. Silvanus Thompson book

Question

In his book Calculas made easy, his example takes this formula: (x+dx) with an exponent of -2 and make it (x exponent-2)(1+dx/x)exponent -2. How does this happen? Shouldn't the exponent -2 turn into exponent -3?

Answer 1

1/(x+dx)²=1/(x²+2xdx+dx²)=1/((x²)(1+2(dx/x)+(dx/x)²)=1/(x²)(1+dx/x)²=x^-2(1+dx/x)^-2

or more directly

1/(x+dx)²=(1/x²)(1/(1+dx/x)²=x^-2(1+dx/x)^-2

now I understand what you are asking, why is f'(x)=-2/x³ when f(x)=1/x²

(f(x+dx)-f(x))/dx=(1/dx)(1/x²)[1/(1+dx/x)²-1]=(1/dx)(1/x²)(-2(dx/x)-(dx/x)²)/(1+(dx/x))²

=(1/dx)(1/x²)(-2(dx/x)-(dx/x)²)(1+2(dx/x)+(dx/x)²)=(1/dx)(1/x²)(-2(dx/x)-4((dx)²/x²)-2(dx/x)³-(dx/x)²-2(dx/x)³-(dx/x)⁴)

=(-2/x³)-(4dx/x⁴)-(4(dx)²/x⁵)-(dx/x⁴)-(dx)³/x⁶)

letting dx approach zero, all terms with dx in numerator go to zero

f'(x)=-2/x³

Answer 2

The transformation from (x + dx)^-2 to x^-2 (1 + dx/x)^-2 is achieved through factoring:

1. Factor out x from the parentheses: (x + dx)^-2 = (x * (1 + dx/x))^-2
2. Apply the exponent rule (ab)^n = a^n * b^n: (x * (1 + dx/x))^-2 = x^-2 * (1 + dx/x)^-2

Why the Exponent Doesn't Change to -3

You're right to be thinking about how exponents change, but in this case, the exponent of -2 applies to the entire term inside the parentheses (x + dx). When we factor out the x, the exponent remains with the factored-out x and also with the remaining expression (1 + dx/x).

Think of it this way:

1. (ab)^n means you are raising the product of a and b to the power of n. This is equivalent to raising a to the power of n and raising b to the power of n, then multiplying the results.

The exponent doesn't change to -3 because we are factoring out a term from the base of the exponentiation, not applying another exponentiation rule to the existing exponent.