Page:Calculus Made Easy.pdf/168

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

${\frac {dx}{dy}}=\epsilon ^{y}$,

and, reverting from the inverse to the original function,

${\frac {dy}{dx}}={\frac {1}{\ {\dfrac {dx}{dy}}\ }}={\frac {1}{\epsilon ^{y}}}={\frac {1}{x}}$.

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

${\frac {d(\log _{\epsilon }x)}{dx}}=x^{-1}$.

Note that $x^{-1}$ 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 $1$. Thus, differentiating $x^{3}$ gave us $3x^{2}$; and differentiating $x^{2}$ gave $2x^{1}$. But differentiating $x^{0}$ does not give us $x^{-1}$ or $0\times x^{-1}$, because $x^{0}$ is itself $=1$, and is a constant. We shall have to come back to this curious fact that differentiating $\log _{\epsilon }x$ gives us ${\dfrac {1}{x}}$ when we reach the chapter on integrating.

 

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Now, try to differentiate

$y=\log _{\epsilon }(x+a)$,

that is

$\epsilon ^{y}=x+a$;

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