Page:Calculus Made Easy.pdf/153

hence

${\frac {dy}{dx}}={\frac {1}{\ {\dfrac {dx}{dy}}\ }}=-{\frac {(1+y^{2})^{2}}{6y}}=-{\frac {\left(1+{\dfrac {3}{x}}-1\right)^{2}}{6\times {\sqrt[{2}]{{\dfrac {3}{x}}-1}}}}=-{\frac {3}{2x^{2}{\sqrt {{\dfrac {3}{x}}-1}}}}$.

Let us take as an other example $y={\dfrac {1}{\sqrt[{3}]{\theta +5}}}$.

The inverse function is $\theta ={\dfrac {1}{y^{3}}}-5$ or $\theta =y^{-3}-5$, and

${\frac {d\theta }{dy}}=-3y^{-4}=-3{\sqrt[{3}]{(\theta +5)^{4}}}$.

It follows that ${\dfrac {dy}{dx}}=-{\dfrac {1}{3{\sqrt {(\theta +5)^{4}}}}}$, as might have been found otherwise.

We shall find this dodge most useful later on; meanwhile you are advised to become familiar with it by verifying by its means the results obtained in Exercises I. (p. 25), Nos. 5, 6, 7; Examples (p. 68), Nos. 1, 2, 4; and Exercises VI. (p. 73), Nos. 1, 2, 3 and 4.

 

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You will surely realize from this chapter and the preceding, that in many respects the calculus is an art rather than a science: an art only to be acquired, as all other arts are, by practice. Hence you should work many examples, and set yourself other examples, to see if you can work them out, until the various artifices become familiar by use.