Page:Calculus Made Easy.pdf/152

Here again

${\frac {dy}{dx}}\times {\frac {dx}{dy}}=1$.

It can be shown that for all functions which can be put into the inverse form, one can always write

${\frac {dy}{dx}}\times {\frac {dx}{dy}}=1\quad$ or $\quad {\frac {dy}{dx}}={\frac {1}{\dfrac {dx}{dy}}}$.

It follows that, being given a function, if it be easier to differentiate the inverse function, this may be done, and the reciprocal of the differential coefficient of the inverse function gives the differential coefficient of the given function itself.

As an example, suppose that we wish to differentiate $y={\sqrt[{2}]{{\dfrac {3}{x}}-1}}$. We have seen one way of doing this, by writing $u={\dfrac {3}{x}}-1$, and finding ${\dfrac {dy}{du}}$ and ${\dfrac {du}{dx}}$ This gives

${\frac {dy}{dx}}=-{\frac {3}{2x^{2}{\sqrt {{\dfrac {3}{x}}-1}}}}$.

If we had forgotten how to proceed by this method, or wished to check our result by some other way of obtaining the differential coefficient, or for any other reason we could not use the ordinary method, we can proceed as follows: The inverse function is $x={\dfrac {3}{1+y^{2}}}$.

${\frac {dx}{dy}}=-{\frac {3\times 2y}{(1+y^{2})^{2}}}=-{\frac {6y}{(1+y^{2})^{2}}}$;