Page:Calculus Made Easy.pdf/219

 a differential coefficient. Secondly, even if it could be differentiated, its differential coefficient (got by slavishly following the usual rule) would be $$0 \times x^{-1}$$, and that multiplication by zero gives it zero value! Therefore when we now come to try to integrate $$x^{-1}\, dx$$, we see that it does not come in anywhere in the powers of $$x$$ that are given by the rule:

It is an exceptional case.

Well; but try again. Look through all the various differentials obtained from various functions of $$x$$, and try to find amongst them $$x^{-1}$$. A sufficient search will show that we actually did get $$\dfrac{dy}{dx} = x^{-1}$$ as the result of differentiating the function $$y = \log_\epsilon x$$ (see p. 148).

Then, of course, since we know that differentiating $$\log_\epsilon x$$ gives us $$x^{-1}$$, we know that, by reversing the process, integrating $$dy = x^{-1}\, dx$$ will give us $$y = \log_\epsilon x$$. But we must not forget the constant factor $$a$$ that was given, nor must we omit to add the undetermined constant of integration. This then gives us as the solution to the present problem,

N.B.–Here note this very remarkable fact, that we could not have integrated in the above case if we had not happened to know the corresponding differentiation. If no one had found out that differentiating $$\log_\epsilon x$$ gave $$x^{-1}$$, we should have been utterly stuck by