C. **Algebraical Functions.** The results of the preceding sections, together with Theorem (6) of § 113, enable us to obtain the derivative of any explicit algebraical function whatsoever.

The most important such function is \(x^{m}\), where \(m\) is a rational number. We have seen already (§ 117) that the derivative of this function is \(mx^{m-1}\) when \(m\) is an integer positive or negative; and we shall now prove that this result is true for all rational values of \(m\). Suppose that \(y = x^{m} = x^{p/q}\), where \(p\) and \(q\) are integers and \(q\) positive; and let \(z = x^{1/q}\), so that \(x = z^{q}\) and \(y = z^{p}\). Then \[\frac{dy}{dx} = \biggl(\frac{dy}{dz}\biggr) \bigg/ \biggl(\frac{dx}{dz}\biggr) = \frac{p}{q} z^{p-q} = mx^{m-1}.\]

This result may also be deduced as a corollary from Ex. XXXVI. 3. For, if \(\phi(x) = x^{m}\), we have \[\begin{aligned} \phi'(x) &= \lim_{h \to 0} \frac{(x + h)^{m} – x^{m}}{h}\\ &= \lim_{\xi \to x} \frac{\xi^{m} – x^{m}}{\xi – x} = mx^{m-1}.\end{aligned}\] It is clear that the more general formula \[\frac{d}{dx} (ax + b)^{m} = ma(ax + b)^{m-1}\] holds also for all rational values of \(m\).

The differentiation of *implicit* algebraical functions involves certain theoretical difficulties to which we shall return in Ch. VII. But there is no practical difficulty in the actual calculation of the derivative of such a function: the method to be adopted will be illustrated sufficiently by an example. Suppose that \(y\) is given by the equation \[x^{3} + y^{3} – 3axy = 0.\] Differentiating with respect to \(x\) we find \[x^{2} + y^{2} \frac{dy}{dx} – a\left(y + x \frac{dy}{dx}\right) = 0\] and so \[\frac{dy}{dx} = -\frac{x^{2} – ay}{y^{2} – ax}.\]

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