118. Differentiation of algebraical functions

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 $m{x}^{m-1}$ 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 $$\frac{dy}{dx}=(\frac{dy}{dz})/(\frac{dx}{dz})=\frac{p}{q}{z}^{p-q}=m{x}^{m-1}.$$

This result may also be deduced as a corollary from Ex. XXXVI. 3. For, if $\varphi (x)=x^m$, we have $$\begin{array}{rl}{\varphi }^{\prime }(x)& =\underset{h\to 0}{lim}\frac{(x+h{)}^m–x^m}{h}\\ & =\underset{\xi \to x}{lim}\frac{{\xi }^m–x^m}{\xi –x}=m{x}^{m-1}.\end{array}$$ 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 $$x^2+y^2\frac{dy}{dx}–a(y+x\frac{dy}{dx})=0$$ and so $$\frac{dy}{dx}=-\frac{x^2–ay}{y^2–ax}.$$