# Alternative to (x+dx)^-2

• Jan 31st 2010, 02:52 AM
p75213
Alternative to (x+dx)^-2
The equation is part of the proof of the derivative of a negative exponent in "Calculus Made Easy" by Silvanus P. Thompson.
The book says (x+dx)^-2 = x^-2(1+dx/x)^-2
I make it x^-2(1+2.dx/x)^-1

I know I am wrong but don't know how.

Note: (dx)^2/x^2 is dropped as it is infintesimal.
• Jan 31st 2010, 03:58 AM
Quote:

Originally Posted by p75213
The equation is part of the proof of the derivative of a negative exponent in "Calculus Made Easy" by Silvanus P. Thompson.
The book says (x+dx)^-2 = x^-2(1+dx/x)^-2
I make it x^-2(1+2.dx/x)^-1

I know I am wrong but don't know how.

Note: (dx)^2/x^2 is dropped as it is infintesimal.

It looks like the book doesn't drop that term after all.

$\displaystyle \frac{1}{x^2+2xdx+(dx)^2}=\frac{1}{x^2}\ \frac{1}{1+\frac{2dx}{x}+\frac{(dx)^2}{x^2}}$

$\displaystyle =\frac{1}{x^2}\ \frac{1}{(1+\frac{dx}{x})^2}$
• Jan 31st 2010, 04:03 AM
CaptainBlack
Quote:

Originally Posted by p75213
The equation is part of the proof of the derivative of a negative exponent in "Calculus Made Easy" by Silvanus P. Thompson.
The book says (x+dx)^-2 = x^-2(1+dx/x)^-2
I make it x^-2(1+2.dx/x)^-1

I know I am wrong but don't know how.

Note: (dx)^2/x^2 is dropped as it is infintesimal.

$\displaystyle (x+y)^2=x^2+2xy+y^2=x^2(1+2y/x + (y/x)^2)=x^2\left(1+\frac{y}{x}\right)^2$

so:

$\displaystyle \frac{1}{(x+dx)^2}=\frac{1}{x^2\left(1+\frac{dx}{x }\right)^2}$

There is no dropping of second degree terms (yet)

CB
• Jan 31st 2010, 11:45 AM
p75213
Thanks Guys,
It was quite simple as it turned out, however I could simply not see the solution at the time.