Sums, Differences, Products and Quotients

Excerpts

Sums, Differences, Products and Quotients
If you have any doubt whether this is right, try a more general case, working it by first principles. And this is the way.
Sums, Differences, Products and Quotients
This justifies the procedure. You differentiate each function separately and add the results.
Sums, Differences, Products and Quotients
The result will certainly *not* be $2x × 4ax^3$; for it is easy to see that neither $c × ax^4$, nor $x^2 × b$, would have been taken into that product.
Sums, Differences, Products and Quotients
*To differentiate the product of two functions, multiply each function by the differential coefficient of the other, and add together the two products so obtained.*
Sums, Differences, Products and Quotients
You should note that this process amounts to the following: Treat $u$ as constant while you differentiate $v$; then treat $v$ as constant while you differentiate $u$; and the whole differential coefficient $\dfrac{dy}{dx}$ will be the sum of these two treatments.
Sums, Differences, Products and Quotients
This gives us our instructions as to *how to differentiate a quotient *of two functions*. Multiply the divisor function by the differential coefficient of the dividend function; then multiply the dividend function by the differential coefficient of the divisor function; and subtract. Lastly divide by the square of the divisor function*.
Sums, Differences, Products and Quotients
The working out of quotients is often tedious, but there is nothing difficult about it.
Sums, Differences, Products and Quotients
Now $du · dv$ is a small quantity of the second order of smallness, and therefore in the limit may be discarded, leaving
Sums, Differences, Products and Quotients
In such a case it is no use to try to work out the division beforehand, because $x^2 + a$ will not divide into $bx^5 + c$, neither have they any common factor.

Equations

Sums, Differences, Products and Quotients
\dfrac{dy}{dx} &= \dfrac{du}{dx} + \dfrac{dv}{dx}
The derivative of a sum of two functions of x is the sum of their derivatives.
Sums, Differences, Products and Quotients
\frac{dy}{dx} &= \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx}
The derivative of a sum of three functions of x is the sum of their three derivatives.
Sums, Differences, Products and Quotients
\frac{dy}{dx} &= \frac{du}{dx} - \frac{dv}{dx}
The derivative of a difference of two functions of x is the difference of their derivatives.
Sums, Differences, Products and Quotients
\dfrac{dy}{dx} = u\, \dfrac{dv}{dx} + v\, \dfrac{du}{dx}
The derivative of a product of two functions is each function times the derivative of the other, added together.
Sums, Differences, Products and Quotients
\dfrac{dy}{dx} &= \dfrac{v\, \dfrac{du}{dx} - u\, \dfrac{dv}{dx}}{v^2}
The derivative of a quotient of two functions is the divisor times the derivative of the dividend, minus the dividend times the derivative of the divisor, all over the square of the divisor.
Sums, Differences, Products and Quotients
V = \dfrac{H}{3} (A + a + \sqrt{Aa} )
The volume of a frustum of a pyramid equals one third of its height times the sum of the two base areas and the square root of their product.
Sums, Differences, Products and Quotients
P = \left( \dfrac{40 + t}{140} \right)^5
Dulong's empirical relation gives the absolute pressure of saturated steam as the fifth power of (40 + t) divided by 140, valid for t above 80 degrees.

Problems

Exercise III

Exercise III, problem 10, p. 47
$y = \dfrac{x^n + a}{x^{-n} + b}$.
Printed answer:
- $\dfrac{anx^{-n-1} + bnx^{n-1} + 2nx^{-1}}{(x^{-n} + b)^2}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes (a*n*x**(-n - 1) + b*n*x**(n - 1) + 2*n*x**(-1))/(x**(-n) + b)**2
Exercise III, problem 11, p. 47
The temperature $t$ of the filament of an incandescent electric lamp is connected to the current passing through the lamp by the relation C = a + bt + ct^2. Find an expression giving the variation of the current corresponding to a variation of temperature.
Printed answer:
- $b + 2ct$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes b + 2*c*t
Exercise III, problem 121, p. 47
The following formulae have been proposed to express the relation between the electric resistance $R$ of a wire at the temperature $t°$C., and the resistance $R_0$ of that same wire at $0°$ Centigrade, $a$, $b$, $c$ being constants. align* R &= R_0(1 + at + bt^2). R &= R_0(1 + at + bt). R &= R_0(1 + at + bt^2)^-1. align* Find the rate of variation of the resistance with regard to temperature as given by each of these formulae.
Printed answer:
- $R_0(a + 2bt)$, $R_0 \left(a + \dfrac{b}{2\sqrt{t}}\right)$, $-\dfrac{R_0(a + 2bt)}{(1 + at + bt^2)^2}$ or $\dfrac{R^2 (a + 2bt)}{R_0}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes R_0*(a + 2*b*t)
Exercise III, problem 122, p. 47
The following formulae have been proposed to express the relation between the electric resistance $R$ of a wire at the temperature $t°$C., and the resistance $R_0$ of that same wire at $0°$ Centigrade, $a$, $b$, $c$ being constants. align* R &= R_0(1 + at + bt^2). R &= R_0(1 + at + bt). R &= R_0(1 + at + bt^2)^-1. align* Find the rate of variation of the resistance with regard to temperature as given by each of these formulae.
Printed answer:
- $R_0(a + 2bt)$, $R_0 \left(a + \dfrac{b}{2\sqrt{t}}\right)$, $-\dfrac{R_0(a + 2bt)}{(1 + at + bt^2)^2}$ or $\dfrac{R^2 (a + 2bt)}{R_0}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes R_0*(a + b/(2*sqrt(t)))
Exercise III, problem 123, p. 47
The following formulae have been proposed to express the relation between the electric resistance $R$ of a wire at the temperature $t°$C., and the resistance $R_0$ of that same wire at $0°$ Centigrade, $a$, $b$, $c$ being constants. align* R &= R_0(1 + at + bt^2). R &= R_0(1 + at + bt). R &= R_0(1 + at + bt^2)^-1. align* Find the rate of variation of the resistance with regard to temperature as given by each of these formulae.
Printed answer:
- $R_0(a + 2bt)$, $R_0 \left(a + \dfrac{b}{2\sqrt{t}}\right)$, $-\dfrac{R_0(a + 2bt)}{(1 + at + bt^2)^2}$ or $\dfrac{R^2 (a + 2bt)}{R_0}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: PASS-ALT-ERRATUM -R_0*(a + 2*b*t)/(1 + a*t + b*t**2)**2
Exercise III, problem 13, p. 47
The electromotive-force $E$ of a certain type of standard cell has been found to vary with the temperature $t$ according to the relation E = 1.4340 [1 - 0.000814(t-15) + 0.000007(t-15)^2] volts. Find the change of electromotive-force per degree, at $15°$, $20°$ and $25°$.
Printed answer:
- $1.4340(0.000014t - \DPtypo{0.000828}{0.001024})$, $-0.00117$, $-0.00107$, $-0.00097$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: PASS-ERRATUM 1.4340*(0.000014*t - 0.001024)
- evaluate: passes 1.4340*(0.000014*t - 0.001024)
- evaluate: passes 1.4340*(0.000014*t - 0.001024)
- evaluate: passes 1.4340*(0.000014*t - 0.001024)
Exercise III, problem 14a, p. 48
The electromotive-force necessary to maintain an electric arc of length $l$ with a current of intensity $i$ has been found by Mrs. Ayrton to be E = a + bl + c + kli, where $a$, $b$, $c$, $k$ are constants. Find an expression for the variation of the electromotiveforce (*a*) with regard to the length of the arc; (*b*) with regard to the strength of the current.
Printed answer:
- $\dfrac{dE}{dl} = b + \dfrac{k}{i}$, $\dfrac{dE}{di} = -\dfrac{c + kl}{i^2}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes b + k/i
Exercise III, problem 14b, p. 48
The electromotive-force necessary to maintain an electric arc of length $l$ with a current of intensity $i$ has been found by Mrs. Ayrton to be E = a + bl + c + kli, where $a$, $b$, $c$, $k$ are constants. Find an expression for the variation of the electromotiveforce (*a*) with regard to the length of the arc; (*b*) with regard to the strength of the current.
Printed answer:
- $\dfrac{dE}{dl} = b + \dfrac{k}{i}$, $\dfrac{dE}{di} = -\dfrac{c + kl}{i^2}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes -(c + k*l)/i**2
Exercise III, problem 1a, p. 46
Differentiateexamples2 SubProbs [(*a*)] $u = 1 + x + \dfrac{x^2}{1 × 2} + \dfrac{x^3}{1 × 2 × 3} + \dotsb$. [(*b*)] $y = ax^2 + bx + c$. (*c*) $y = (x + a)^2$. [(*d*)] $y = (x + a)^3$. SubProbs
Printed answer:
- (*a*) $1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \dfrac{x^4}{24} + \ldots$ (*b*) $2ax + b$. (*c*) $2x + 2a$. (*d*) $3x^2 + 6ax + 3a^2$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes, with the problem read into an equation exp(x)
Exercise III, problem 1b, p. 46
Differentiateexamples2 SubProbs [(*a*)] $u = 1 + x + \dfrac{x^2}{1 × 2} + \dfrac{x^3}{1 × 2 × 3} + \dotsb$. [(*b*)] $y = ax^2 + bx + c$. (*c*) $y = (x + a)^2$. [(*d*)] $y = (x + a)^3$. SubProbs
Printed answer:
- (*a*) $1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \dfrac{x^4}{24} + \ldots$ (*b*) $2ax + b$. (*c*) $2x + 2a$. (*d*) $3x^2 + 6ax + 3a^2$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 2*a*x + b
Exercise III, problem 1c, p. 46
Differentiateexamples2 SubProbs [(*a*)] $u = 1 + x + \dfrac{x^2}{1 × 2} + \dfrac{x^3}{1 × 2 × 3} + \dotsb$. [(*b*)] $y = ax^2 + bx + c$. (*c*) $y = (x + a)^2$. [(*d*)] $y = (x + a)^3$. SubProbs
Printed answer:
- (*a*) $1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \dfrac{x^4}{24} + \ldots$ (*b*) $2ax + b$. (*c*) $2x + 2a$. (*d*) $3x^2 + 6ax + 3a^2$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 2*x + 2*a
Exercise III, problem 1d, p. 46
Differentiateexamples2 SubProbs [(*a*)] $u = 1 + x + \dfrac{x^2}{1 × 2} + \dfrac{x^3}{1 × 2 × 3} + \dotsb$. [(*b*)] $y = ax^2 + bx + c$. (*c*) $y = (x + a)^2$. [(*d*)] $y = (x + a)^3$. SubProbs
Printed answer:
- (*a*) $1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \dfrac{x^4}{24} + \ldots$ (*b*) $2ax + b$. (*c*) $2x + 2a$. (*d*) $3x^2 + 6ax + 3a^2$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 3*x**2 + 6*a*x + 3*a**2
Exercise III, problem 2, p. 46
If $w = at - \frac{1}{2}bt^2$, find $\dfrac{dw}{dt}$.
Printed answer:
- $\dfrac{dw}{dt} = a - bt$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes a - b*t
Exercise III, problem 3, p. 46
Find the differential coefficient of y = (x + -1) × (x - -1).
Printed answer:
- $\dfrac{dy}{dx} = 2x$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 2*x
Exercise III, problem 4, p. 46
Differentiate y = (197x - 34x^2) × (7 + 22x - 83x^3).
Printed answer:
- $14110x^4 - 65404x^3 - 2244x^2 + 8192x + 1379$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 14110*x**4 - 65404*x**3 - 2244*x**2 + 8192*x + 1379
Exercise III, problem 5, p. 46
If $x = (y + 3) × (y + 5)$, find $\dfrac{dx}{dy}$.
Printed answer:
- $\dfrac{dx}{dy} = 2y + 8$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 2*y + 8
Exercise III, problem 6, p. 46
Differentiate $y = 1.3709x × (112.6 + 45.202x^2)$.
Printed answer:
- $185.9022654x^2 + 154.36334$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 185.9022654*x**2 + 154.36334
Exercise III, problem 7, p. 47
$y = \dfrac{2x + 3}{3x + 2}$.
Printed answer:
- $\dfrac{-5}{(3x + 2)^2}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes -5/(3*x + 2)**2
Exercise III, problem 8, p. 47
$y = \dfrac{1 + x + 2x^2 + 3x^3}{1 + x + 2x^2}$.
Printed answer:
- $\dfrac{6x^4 + 6x^3 + 9x^2}{(1 + x + 2x^2)^2}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes (6*x**4 + 6*x**3 + 9*x**2)/(1 + x + 2*x**2)**2
Exercise III, problem 9, p. 47
$y = \dfrac{ax + b}{cx + d}$.
Printed answer:
- $\dfrac{ad - bc}{(cx + d)^2}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes (a*d - b*c)/(c*x + d)**2