Other Useful Dodges
Excerpts
Other Useful Dodges
If we perform many additions of two or more fractions the denominators of which contain only terms in $x$, and no terms in $x^2$, $x^3$, or any other powers of $x$, we *always* find that *the denominator of the final resulting fraction is the product of the denominators* of the fractions which were added to form the result.
Other Useful Dodges
If we make $x=1$, we get $4 = (A × 0)+(B × 2)$, so that $B=2$; and if we make $x=-1$, we get $-2 = (A × -2) + (B × 0)$, so that $A=1$.
Other Useful Dodges
Since the given fraction and the fraction found by adding the partial fractions are equal, and have *identical* denominators, the numerators must also be identically the same. In such a case, and for such algebraical expressions as those with which we are dealing here, *the coefficients of the same powers of $x$ are equal and of same sign*.
Other Useful Dodges
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.
Other Useful Dodges
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.
Other Useful Dodges
The equation must be true for all values of $x$; therefore it must be true for such values of $x$ as will cause $x-1$ and $x+1$ to become zero, that is for $x=1$ and for $x=-1$ respectively.
Other Useful Dodges
We see that it is sufficient to allow for one numerical term in each numerator, and that we always get the ultimate partial fractions.
Other Useful Dodges
It is useful to check the results obtained. The simplest way is to replace $x$ by a single value, say $+1$, both in the given expression and in the partial fractions obtained.
Other Useful Dodges
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.
Other Useful Dodges
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.
Equations
Other Useful Dodges
\frac{3}{x+1} - \frac{1}{x-1} + \frac{2}{x+3}The fraction (4x^2+2x-14)/(x^3+3x^2-x-3) splits into three partial fractions with denominators x+1, x-1 and x+3.
Other Useful Dodges
\frac{x-1}{x^2+1} - \frac{2}{x+1}The fraction (-x^2-3)/((x^2+1)(x+1)) splits into (x-1)/(x^2+1) minus 2/(x+1).
Other Useful Dodges
\frac{2}{x+1} - \frac{2}{(x+1)^2} + \frac{1}{x-2}The fraction (3x^2-2x+1)/((x+1)^2(x-2)) splits into partial fractions with denominators x+1, (x+1)^2 and x-2.
Other Useful Dodges
\frac{(8x - 5)}{(2x^2 - 1)^2} + \frac{8(x - 1)}{2x^2 - 1} - \frac{4}{x + 1}The fraction (3x-1)/((2x^2-1)^2(x+1)) splits into (8x-5)/(2x^2-1)^2, 8(x-1)/(2x^2-1) and -4/(x+1).
Other Useful Dodges
\frac{4}{(x + 1)^2} - \frac{3}{(x + 1)^3}The fraction (4x+1)/(x+1)^3 splits into 4/(x+1)^2 minus 3/(x+1)^3, found by substituting z = x+1.
Other Useful Dodges
\frac{3x^2 - 2x + 1}{(x+1)^2(x-2)} = \frac{x-1}{(x+1)^2} + \frac{1}{x+1} + \frac{1}{x-2}The fraction (3x^2-2x+1)/((x+1)^2(x-2)) equals the three-term sum with numerator x-1 over (x+1)^2, which is not the form that the method of unknown numerators yields.
Other Useful Dodges
\frac{dy}{dx} = -\frac{3}{(3x-1)^2} + \frac{4}{(2x+3)^2}The derivative of y = (5-4x)/(6x^2+7x-3), obtained by differentiating its partial fractions 1/(3x-1) - 2/(2x+3).
Other Useful Dodges
\frac{dy}{dx} = -\frac{3}{2x^2\sqrt{\dfrac{3}{x} -1}}The derivative of y = sqrt(3/x - 1), found by differentiating the inverse function x = 3/(1+y^2).
Other Useful Dodges
\dfrac{dy}{dx} = -\dfrac{1}{3\sqrt{(\theta +5)^4}}The derivative of y = 1/cuberoot(theta+5) with respect to x, obtained from the inverse function theta = y^(-3) - 5.
Other Useful Dodges
\frac{dy}{dx} × \frac{dx}{dy} = 1The derivative of y with respect to x times the derivative of x with respect to y equals 1, illustrated for y = 3x and y = 4x^2.
Other Useful Dodges
\frac{dy}{dx} = \frac{1}{\ \dfrac{dx}{dy}\ }For any function that has an inverse form, the derivative of the function is the reciprocal of the derivative of its inverse function.
Problems
Exercise XI
Exercise XI, problem 1, p. 130
$\dfrac{3x + 5}{(x - 3)(x + 4)}$.
Printed answer:- $\dfrac{2}{ x - 3} + \dfrac{1}{ x + 4}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem2/(x-3)+1/(x+4)
Exercise XI, problem 10, p. 131
$\dfrac{x^4 + 1}{x^3 + 1}$.
Printed answer:- $x + \dfrac{2}{3(x + 1)} + \dfrac{1 - 2x}{3(x^2 - x + 1)}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problemx+2/(3*(x+1))+(1-2*x)/(3*(x**2-x+1))
Exercise XI, problem 11, p. 131
$\dfrac{5x^2 + 6x + 4}{(x +1)(x^2 + x + 1)}$.
Printed answer:- $\dfrac{3}{(x + 1)} + \dfrac{2x + 1}{x^2 + x + 1}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem3/(x+1)+(2*x+1)/(x**2+x+1)
Exercise XI, problem 12, p. 131
$\dfrac{x}{(x - 1)(x - 2)^2}$.
Printed answer:- $\dfrac{1}{ x - 1} - \dfrac{1}{ x - 2} + \dfrac{2}{(x - 2)^2}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem1/(x-1)-1/(x-2)+2/(x-2)**2
Exercise XI, problem 13, p. 131
$\dfrac{x}{(x^2 - 1)(x + 1)}$.
Printed answer:- $\dfrac{1}{4(x - 1)} - \dfrac{1}{4(x + 1)} + \dfrac{1}{2(x + 1)^2}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem1/(4*(x-1))-1/(4*(x+1))+1/(2*(x+1)**2)
Exercise XI, problem 14, p. 131
$\dfrac{x + 3}{ (x +2)^2(x - 1)}$.
Printed answer:- $\dfrac{4}{9(x - 1)} - \dfrac{4}{9(x + 2)} - \dfrac{1}{3(x + 2)^2}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem4/(9*(x-1))-4/(9*(x+2))-1/(3*(x+2)**2)
Exercise XI, problem 15, p. 131
$\dfrac{3x^2 + 2x + 1}{(x + 2)(x^2 + x + 1)^2}$.
Printed answer:- $\dfrac{1}{ x + 2} - \dfrac{x - 1}{ x^2 + x + 1} - \dfrac{1}{(x^2 + x + 1)^2}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem1/(x+2)-(x-1)/(x**2+x+1)-1/(x**2+x+1)**2
Exercise XI, problem 16, p. 131
$\dfrac{5x^2 + 8x - 12}{(x + 4)^3}$.
Printed answer:- $\dfrac{5}{ x + 4} -\dfrac{32}{(x + 4)^2} + \dfrac{36}{(x + 4)^3}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem5/(x+4)-32/(x+4)**2+36/(x+4)**3
Exercise XI, problem 17, p. 131
$\dfrac{7x^2 + 9x - 1}{(3x - 2)^4}$.
Printed answer:- $\dfrac{7}{9(3x - 2)^2} + \dfrac{55}{9(3x - 2)^3} + \dfrac{73}{9(3x - 2)^4}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem7/(9*(3*x-2)**2)+55/(9*(3*x-2)**3)+73/(9*(3*x-2)**4)
Exercise XI, problem 18, p. 131
$\dfrac{x^2}{(x^3 - 8)(x - 2)}$.
Printed answer:- $\dfrac{1}{6(x - 2)} + \dfrac{1}{3(x - 2)^2} - \dfrac{x}{6(x^2 + 2x + 4)}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem1/(6*(x-2))+1/(3*(x-2)**2)-x/(6*(x**2+2*x+4))
Exercise XI, problem 2, p. 130
$\dfrac{3x - 4}{(x - 1)(x - 2)}$.
Printed answer:- $\dfrac{1}{ x - 1} + \dfrac{2}{ x - 2}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem1/(x-1)+2/(x-2)
Exercise XI, problem 3, p. 130
$\dfrac{3x + 5}{x^2 + x - 12}$.
Printed answer:- $\dfrac{2}{ x - 3} + \dfrac{1}{ x + 4}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem2/(x-3)+1/(x+4)
Exercise XI, problem 4, p. 130
$\dfrac{x + 1}{x^2 - 7x + 12}$.
Printed answer:- $\dfrac{5}{ x - 4} - \dfrac{4}{ x - 3}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem5/(x-4)-4/(x-3)
Exercise XI, problem 5, p. 130
$\dfrac{x - 8}{(2x + 3)(3x - 2)}$.
Printed answer:- $\dfrac{19}{13(2x + 3)} - \dfrac{22}{13(3x - 2)}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem19/(13*(2*x+3))-22/(13*(3*x-2))
Exercise XI, problem 6, p. 130
$\dfrac{x^2 - 13x + 26}{(x - 2)(x - 3)(x - 4)}$.
Printed answer:- $\dfrac{2}{ x - 2} + \dfrac{4}{ x - 3} - \dfrac{5}{ x - 4}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem2/(x-2)+4/(x-3)-5/(x-4)
Exercise XI, problem 7, p. 131
$\dfrac{x^2 - 3x + 1}{(x - 1)(x + 2)(x - 3)}$.
Printed answer:- $\dfrac{1}{6(x - 1)} + \dfrac{11}{15(x + 2)} + \dfrac{1}{10(x - 3)}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem1/(6*(x-1))+11/(15*(x+2))+1/(10*(x-3))
Exercise XI, problem 8, p. 131
$\dfrac{5x^2 + 7x + 1}{(2x + 1)(3x - 2)(3x + 1)}$.
Printed answer:- $\dfrac{7}{9(3x + 1)} + \dfrac{71}{63(3x - 2)} - \dfrac{5}{7(2x + 1)}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem7/(9*(3*x+1))+71/(63*(3*x-2))-5/(7*(2*x+1))
Exercise XI, problem 9, p. 131
$\dfrac{x^2}{x^3 - 1}$.
Printed answer:- $\dfrac{1}{3(x - 1)} + \dfrac{2x + 1}{3(x^2 + x + 1)}$.
unverified: no computed check settled this one (yet)
How it was checked
other: the check does not fit this problem1/(3*(x-1))+(2*x+1)/(3*(x**2+x+1))