113. General rules for differentiation

Throughout the theorems which follow we assume that the functions $f(x)$ and $F(x)$ have derivatives $f^{\prime }(x)$ and $F^{\prime }(x)$ for the values of $x$ considered.

(1) If $\varphi (x)=f(x)+F(x)$, then $\varphi (x)$ has a derivative $${\varphi }^{\prime }(x)=f^{\prime }(x)+F^{\prime }(x).$$

(2) If $\varphi (x)=kf(x)$, where $k$ is a constant, then $\varphi (x)$ has a derivative $${\varphi }^{\prime }(x)=k{f}^{\prime }(x).$$

We leave it as an exercise to the reader to deduce these results from the general theorems stated in Ex. XXXV. 1.

(3) If $\varphi (x)=f(x)F(x)$, then $\varphi (x)$ has a derivative $${\varphi }^{\prime }(x)=f(x)F^{\prime }(x)+f^{\prime }(x)F(x).$$

For $$\begin{array}{rl}{\varphi }^{\prime }(x)& =lim\frac{f(x+h)F(x+h)–f(x)F(x)}{h}\\ & =lim\{f(x+h)\frac{F(x+h)–F(x)}{h}+F(x)\frac{f(x+h)–f(x)}{h}\}\\ & =f(x)F^{\prime }(x)+F(x)f^{\prime }(x).\end{array}$$

(4) If $\varphi (x)={\displaystyle \frac{1}{f(x)}}$, then $\varphi (x)$ has a derivative $${\varphi }^{\prime }(x)=-\frac{f^{\prime }(x)}{\{f(x){\}}^2}.$$

In this theorem we of course suppose that $f(x)$ is not equal to zero for the particular value of $x$ under consideration. Then $${\varphi }^{\prime }(x)=lim\frac{1}{h}\left\{\frac{f(x)–f(x+h)}{f(x+h)f(x)}\right\}=-\frac{f^{\prime }(x)}{\{f(x){\}}^2}.$$

(5) If $\varphi (x)={\displaystyle \frac{f(x)}{F(x)}}$, then $\varphi (x)$ has a derivative $${\varphi }^{\prime }(x)=\frac{f^{\prime }(x)F(x)–f(x)F^{\prime }(x)}{\{F(x){\}}^2}.$$

This follows at once from (3) and (4).

(6) If $\varphi (x)=F\{f(x)\}$, then $\varphi (x)$ has a derivative $${\varphi }^{\prime }(x)=F^{\prime }\{f(x)\}{f}^{\prime }(x).$$

For let $$f(x)=y,f(x+h)=y+k.$$ Then $k\to 0$ as $h\to 0$, and $k/h\to f^{\prime }(x)$. And $$\begin{array}{rl}{\varphi }^{\prime }(x)& =lim\frac{F\{f(x+h)\}–F\{f(x)\}}{h}\\ & =lim\left\{\frac{F(y+k)–F(y)}{k}\right\}\times lim\left(\frac{k}{h}\right)\\ & =F^{\prime }(y)f^{\prime }(x).\end{array}$$

This theorem includes (2) and (4) as special cases, as we see on taking $F(x)=kx$ or $F(x)=1/x$. Another interesting special case is that in which $f(x)=ax+b$: the theorem then shows that the derivative of $F(ax+b)$ is $a{F}^{\prime }(ax+b)$.

Our last theorem requires a few words of preliminary explanation. Suppose that $x=\psi (y)$, where $\psi (y)$ is continuous and steadily increasing (or decreasing), in the stricter sense of § 95, in a certain interval of values of $y$. Then we may write $y=\varphi (x)$, where $\varphi$ is the ‘inverse’ function (§ 109) of $\psi$.

(7) If $y=\varphi (x)$, where $\varphi$ is the inverse function of $\psi$, so that $x=\psi (y)$, and $\psi (y)$ has a derivative ${\psi }^{\prime }(y)$ which is not equal to zero, then $\varphi (x)$ has a derivative $${\varphi }^{\prime }(x)=\frac{1}{{\psi }^{\prime }(y)}.$$

For if $\varphi (x+h)=y+k$, then $k\to 0$ as $h\to 0$, and $${\varphi }^{\prime }(x)=\underset{h\to 0}{lim}\frac{\varphi (x+h)–\varphi (x)}{(x+h)–x}=\underset{k\to 0}{lim}\frac{(y+k)–y}{\psi (y+k)–\psi (y)}=\frac{1}{{\psi }^{\prime }(y)}.$$ The last function may now be expressed in terms of $x$ by means of the relation $y=\varphi (x)$, so that ${\varphi }^{\prime }(x)$ is the reciprocal of ${\psi }^{\prime }\{\varphi (x)\}$. This theorem enables us to differentiate any function if we know the derivative of the inverse function.

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$\leftarrow$ 110-112. Derivatives

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114. Derivatives of complex functions $\to$