Page:Calculus Made Easy.pdf/44

Expanding this by the binomial theorem (see p. 141), we get

$=x^{-2}\left[1-{\frac {2dx}{x}}+{\frac {2(2+1)}{1\times 2}}{\left({\frac {dx}{x}}\right)}^{2}-etc.\right]$

$=x^{-2}-2x^{-3}\cdot dx+3x^{-4}(dx)^{2}-4x^{-5}(dx)^{3}+etc$.

So, neglecting the small quantities of higher orders of smallness, we have:

$y+dy=x^{-2}-2x^{-3}\cdot dx$.

Subtracting the original $y=x^{-2}$, we find

$dy\,\,=-2x^{-3}dx$,

${\frac {dy}{dx}}=-2x^{-3}$.

And this is still in accordance with the rule inferred above.

Case of a fractional power.

Let $y=x^{\tfrac {1}{2}}$. Then, as before,

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

$={\sqrt {x}}+{\frac {1}{2}}{\frac {dx}{\sqrt {x}}}-{\frac {1}{8}}{\frac {(dx)^{2}}{x{\sqrt {x}}}}+$ terms with higher powers of $dx$.

Subtracting the original $y=x^{\tfrac {1}{2}}$, and neglecting higher powers we have left:

$dy={\frac {1}{2}}{\frac {dx}{\sqrt {x}}}={\frac {1}{2}}x^{-{\tfrac {1}{2}}}\cdot dx$,

and ${\frac {dy}{dx}}={\frac {1}{2}}x^{-{\tfrac {1}{2}}}$. Agreeing with the general rule.