Page:Calculus Made Easy.pdf/168

whence, since the differential of ${\displaystyle \epsilon ^{y}}$ with regard to ${\displaystyle y}$ is the original function unchanged (see p. 143),

and, reverting from the inverse to the original function,

Now this is a very curious result. It may be written

Note that ${\displaystyle x^{-1}}$ is a result that we could never have got by the rule for differentiating powers. That rule (page 25) is to multiply by the power, and reduce the power by ${\displaystyle 1}$. Thus, differentiating ${\displaystyle x^{3}}$ gave us ${\displaystyle 3x^{2}}$; and differentiating ${\displaystyle x^{2}}$ gave ${\displaystyle 2x^{1}}$. But differentiating ${\displaystyle x^{0}}$ does not give us ${\displaystyle x^{-1}}$ or ${\displaystyle 0\times x^{-1}}$, because ${\displaystyle x^{0}}$ is itself ${\displaystyle =1}$, and is a constant. We shall have to come back to this curious fact that differentiating ${\displaystyle \log _{\epsilon }x}$ gives us ${\displaystyle {\dfrac {1}{x}}}$ when we reach the chapter on integrating.

Now, try to differentiate

that is

we have ${\displaystyle {\dfrac {d(x+a)}{dy}}=\epsilon ^{y}}$, since the differential of ${\displaystyle \epsilon ^{y}}$ remains ${\displaystyle \epsilon ^{y}}$.