Page:Calculus Made Easy.pdf/44

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

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.