Throughout the theorems which follow we assume that the functions \(f(x)\) and \(F(x)\) have derivatives \(f'(x)\) and \(F'(x)\) for the values of \(x\) considered.

(1) If \(\phi(x) = f(x) + F(x)\), then \(\phi(x)\) has a derivative \[\phi'(x) = f'(x) + F'(x).\]

(2) If \(\phi(x) = kf(x)\), where \(k\) is a constant, then \(\phi(x)\) has a derivative \[\phi'(x) = kf'(x).\]

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

(3) If \(\phi(x) = f(x)F(x)\), then \(\phi(x)\) has a derivative \[\phi'(x) = f(x)F'(x) + f'(x)F(x).\]

For \[\begin{aligned} \phi'(x) &= \lim\frac{f(x + h)F(x + h) – f(x)F(x)}{h}\\ &= \lim\left\{f(x + h)\frac{F(x + h) – F(x)}{h} + F(x)\frac{f(x + h) – f(x)}{h}\right\}\\ &=f(x)F'(x) + F(x)f'(x).\end{aligned}\]

(4) If \(\phi(x) = \dfrac{1}{f(x)}\), then \(\phi(x)\) has a derivative \[\phi'(x) = -\frac{f'(x)}{\{f(x)\}^{2}}.\]

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

(5) If \(\phi(x) = \dfrac{f(x)}{F(x)}\), then \(\phi(x)\) has a derivative \[\phi'(x) = \frac{f'(x)F(x) – f(x)F'(x)}{\{F(x)\}^{2}}.\]

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

(6) If \(\phi(x) = F\{f(x)\}\), then \(\phi(x)\) has a derivative \[\phi'(x) = F’\{f(x)\} f'(x).\]

For let \[f(x) = y,\quad f(x + h) = y + k.\] Then \(k \to 0\) as \(h \to 0\), and \(k/h \to f'(x)\). And \[\begin{aligned} \phi'(x) & = \lim \frac{F\{f(x + h)\} – F\{f(x)\}}{h}\\ & = \lim \left\{\frac{F(y + k) – F(y)}{k}\right\} \times \lim \left(\frac{k}{h}\right)\\ & = F'(y)f'(x).\end{aligned}\]

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 \(aF'(ax + b)\).

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 \(y = \phi(x)\), where \(\phi\) is the ‘inverse’ function (§ 109) of \(\psi\).

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

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

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