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),

and, reverting from the inverse to the original function,

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

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.

Now, try to differentiate

that is

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