Other Useful Dodges

We have seen that when we differentiate a fraction we have to perform a rather complicated operation; and, if the fraction is not itself a simple one, the result is bound to be a complicated expression. If we could split the fraction into two or more simpler fractions such that their sum is equivalent to the original fraction, we could then proceed by differentiating each of these simpler expressions. And the result of differentiating would be the sum of two (or more) differentials, each one of which is relatively simple; while the final expression, though of course it will be the same as that which could be obtained without resorting to this dodge, is thus obtained with much less effort and appears in a simplified form.

Let us see how to reach this result. Try first the
job of adding two fractions together to form a resultant
fraction. Take, for example, the two fractions $\dfrac{1}{x+1}$
and $\dfrac{2}{x-1}$. Every schoolboy can add these together
and find their sum to be $\dfrac{3x+1}{x^2-1}$. And in the same
way he can add together three or more fractions.
Now this process can certainly be reversed: that is to
say, that if this last expression were given, it is certain
that it can somehow be split back again into its
original components or partial fractions. Only we do
not know in every case that may be presented to us
*how* we can so split it. In order to find this out
we shall consider a simple case at first. But it is
important to bear in mind that all which follows
applies only to what are called “proper” algebraic
fractions, meaning fractions like the above, which have
the numerator of *a lesser degree* than the denominator;
that is, those in which the highest index of $x$ is less
in the numerator than in the denominator. If we
have to deal with such an expression as $\dfrac{x^2+2}{x^2-1}$, we can
simplify it by division, since it is equivalent to
$1+\dfrac{3}{x^2-1}$; and $\dfrac{3}{x^2-1}$ is a proper algebraic fraction
to which the operation of splitting into partial fractions
can be applied, as explained hereafter.

Proceeding as before, we find \[ 4x^2 + 2x - 14 = A(x-1)(x+3) + B(x+1)(x+3) + C(x+1)(x-1). \]

If $x= -1$, we get: \[ -12 = A(-2 × 2) + (B × 0) + (C × 0);\quad \text{whence } A = 3. \]

If $x = -3$, we get: \[ 16 = (A × 0) + (B × 0) + C(-2 × -4);\quad \text{whence } C = 2. \]

Take as another example the fraction \[ \frac{x^3-2}{(x^2+1)(x^2+2)}. \]

Whenever the denominator contains but a power of a single factor, a very quick method is as follows:

Taking, for example, $\dfrac{4x + 1}{(x + 1)^3}$, let $x + 1 = z$; then $x = z - 1$.

The partial fractions are, therefore, \[ \frac{4}{(x + 1)^2} - \frac{3}{(x + 1)^3}. \]

If we split the given expression into
\[
\frac{1}{3x-1} - \frac{2}{2x+3},
\]
we get, however,
\[
\frac{dy}{dx} = -\frac{3}{(3x-1)^2} + \frac{4}{(2x+3)^2},
\]
which is really the same result as above split into
partial fractions. But the splitting, if done after
differentiating, is more complicated, as will easily be
seen. When we shall deal with the *integration* of
such expressions, we shall find the splitting into
partial fractions a precious auxiliary (see here).

Split into fractions:

(1) $\dfrac{3x + 5}{(x - 3)(x + 4)}$.

(2) $\dfrac{3x - 4}{(x - 1)(x - 2)}$.

(3) $\dfrac{3x + 5}{x^2 + x - 12}$.

(4) $\dfrac{x + 1}{x^2 - 7x + 12}$.

(5) $\dfrac{x - 8}{(2x + 3)(3x - 2)}$.

(6) $\dfrac{x^2 - 13x + 26}{(x - 2)(x - 3)(x - 4)}$.

(7) $\dfrac{x^2 - 3x + 1}{(x - 1)(x + 2)(x - 3)}$.

(8) $\dfrac{5x^2 + 7x + 1}{(2x + 1)(3x - 2)(3x + 1)}$.

(9) $\dfrac{x^2}{x^3 - 1}$.

(10) $\dfrac{x^4 + 1}{x^3 + 1}$.

(11) $\dfrac{5x^2 + 6x + 4}{(x +1)(x^2 + x + 1)}$.

(12) $\dfrac{x}{(x - 1)(x - 2)^2}$.

(13) $\dfrac{x}{(x^2 - 1)(x + 1)}$.

(14) $\dfrac{x + 3}{ (x +2)^2(x - 1)}$.

(15) $\dfrac{3x^2 + 2x + 1}{(x + 2)(x^2 + x + 1)^2}$.

(16) $\dfrac{5x^2 + 8x - 12}{(x + 4)^3}$.

(17) $\dfrac{7x^2 + 9x - 1}{(3x - 2)^4}$.

(18) $\dfrac{x^2}{(x^3 - 8)(x - 2)}$.

(1) $\dfrac{2}{ x - 3} + \dfrac{1}{ x + 4}$.

(2) $\dfrac{1}{ x - 1} + \dfrac{2}{ x - 2}$.

(3) $\dfrac{2}{ x - 3} + \dfrac{1}{ x + 4}$.

(4) $\dfrac{5}{ x - 4} - \dfrac{4}{ x - 3}$.

(5) $\dfrac{19}{13(2x + 3)} - \dfrac{22}{13(3x - 2)}$.

(6) $\dfrac{2}{ x - 2} + \dfrac{4}{ x - 3} - \dfrac{5}{ x - 4}$.

(7) $\dfrac{1}{6(x - 1)} + \dfrac{11}{15(x + 2)} + \dfrac{1}{10(x - 3)}$.

(8) $\dfrac{7}{9(3x + 1)} + \dfrac{71}{63(3x - 2)} - \dfrac{5}{7(2x + 1)}$.

(9) $\dfrac{1}{3(x - 1)} + \dfrac{2x + 1}{3(x^2 + x + 1)}$.

(10) $x + \dfrac{2}{3(x + 1)} + \dfrac{1 - 2x}{3(x^2 - x + 1)}$.

(11) $\dfrac{3}{(x + 1)} + \dfrac{2x + 1}{x^2 + x + 1}$.

(12) $\dfrac{1}{ x - 1} - \dfrac{1}{ x - 2} + \dfrac{2}{(x - 2)^2}$.

(13) $\dfrac{1}{4(x - 1)} - \dfrac{1}{4(x + 1)} + \dfrac{1}{2(x + 1)^2}$.

(14) $\dfrac{4}{9(x - 1)} - \dfrac{4}{9(x + 2)} - \dfrac{1}{3(x + 2)^2}$.

(15) $\dfrac{1}{ x + 2} - \dfrac{x - 1}{ x^2 + x + 1} - \dfrac{1}{(x^2 + x + 1)^2}$.

(16) $\dfrac{5}{ x + 4} -\dfrac{32}{(x + 4)^2} + \dfrac{36}{(x + 4)^3}$.

(17) $\dfrac{7}{9(3x - 2)^2} + \dfrac{55}{9(3x - 2)^3} + \dfrac{73}{9(3x - 2)^4}$.

(18) $\dfrac{1}{6(x - 2)} + \dfrac{1}{3(x - 2)^2} - \dfrac{x}{6(x^2 + 2x + 4)}$.

Consider the function (see here) $y = 3x$; it can be
expressed in the form $x = \dfrac{y}{3}$; this latter form is called
the *inverse function* to the one originally given.

If $y = 3x$, $\dfrac{dy}{dx} = 3$; if $x=\dfrac{y}{3}$, $\dfrac{dx}{dy} = \dfrac{1}{3}$, and we see that \[ \frac{dy}{dx} = \frac{1}{\ \dfrac{dx}{dy}\ }\quad \text{or}\quad \frac{dy}{dx} × \frac{dx}{dy} = 1. \]

Consider $y= 4x^2$, $\dfrac{dy}{dx} = 8x$; the inverse function is \[ x = \frac{y^{\frac{1}{2}}}{2},\quad \text{and}\quad \frac{dx}{dy} = \frac{1}{4\sqrt{y}} = \frac{1}{4 × 2x} = \frac{1}{8x}. \] \begin{align*} \text{Here again}\; \frac{dy}{dx}×\frac{dx}{dy} &= 1. \end{align*}

It can be shown that for all functions which can be put into the inverse form, one can always write \[ \frac{dy}{dx} × \frac{dx}{dy} = 1\quad \text{or}\quad \frac{dy}{dx} = \frac{1}{\ \dfrac{dx}{dy}\ }. \]

It follows that, being given a function, if it be easier to differentiate the inverse function, this may be done, and the reciprocal of the differential coefficient of the inverse function gives the differential coefficient of the given function itself.

As an example, suppose that we wish to differentiate $y=\sqrt[2]{\dfrac{3}{x}-1}$. We have seen one way of doing this, by writing $u=\dfrac{3}{x}-1$, and finding $\dfrac{dy}{du}$ and $\dfrac{du}{dx}$. This gives \[ \frac{dy}{dx} = -\frac{3}{2x^2\sqrt{\dfrac{3}{x} -1}}. \]

If we had forgotten how to proceed by this method, or wished to check our result by some other way of obtaining the differential coefficient, or for any other reason we could not use the ordinary method, we can proceed as follows: The inverse function is $x=\dfrac{3}{1+y^2}$. \[ \frac{dx}{dy} = -\frac{3 × 2y}{(1+y^2)^2} = -\frac{6y}{(1+y^2)^2}; \] hence \[ \frac{dy}{dx} = \frac{1}{\ \dfrac{dx}{dy}\ } = -\frac{(1+y^2)^2}{6y} = -\frac{\left(1+\dfrac{3}{x} -1\right)^2}{6×\sqrt[2]{\dfrac{3}{x}-1}} = -\frac{3}{2x^2\sqrt{\dfrac{3}{x}-1}}. \]

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. (here), Nos. 5, 6, 7; Examples (here), Nos. 1, 2, 4; and Exercises VI. (here), Nos. 1, 2, 3 and 4.

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.

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