# Calculus Made Easy/Chapter 6

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CHAPTER VI.
SUMS, DIFFERENCES, PRODUCTS AND QUOTIENTS.

We have learned how to differentiate simple algebraical functions such as ${\displaystyle x^{2}+c}$ or ${\displaystyle ax^{4}}$, and we have now to consider how to tackle the sum of two or more functions.

For instance, let

${\displaystyle y=(x^{2}+c)+(ax^{4}+b)}$;

what will its ${\displaystyle {\frac {dy}{dx}}}$ be? How are we to go to work on this new job?

The answer to this question is quite simple: just differentiate them, one after the other, thus:

${\displaystyle {\frac {dy}{dx}}=2x+4ax^{3}}$. (Ans.)

If you have any doubt whether this is right, try a more general case, working it by first principles. And this is the way.

Let ${\displaystyle y=u+v}$, where u is any function of ${\displaystyle x}$, and ${\displaystyle v}$ any other function of ${\displaystyle x}$. Then, letting ${\displaystyle x}$ increase to ${\displaystyle x+dx}$, ${\displaystyle y}$ will increase to ${\displaystyle y+dy}$; and ${\displaystyle u}$ will increase to ${\displaystyle u+du}$; and ${\displaystyle v}$ to ${\displaystyle v+dv}$.

And we shall have:

${\displaystyle y+dy=u+du+v+dv}$.

Subtracting the original ${\displaystyle y=u+v}$, we get

${\displaystyle dy=du+dv}$,

and dividing through by ${\displaystyle dx}$, we get:

${\displaystyle {\frac {dy}{dx}}={\frac {du}{dx}}+{\frac {dv}{dx}}}$.

This justifies the procedure. You differentiate each function separately and add the results. So if now we take the example of the preceding paragraph, and put in the values of the two functions, we shall have, using the notation shown (p. 17),

${\displaystyle {\frac {dy}{dx}}={\frac {d(x^{2}+c)}{dx}}+{\frac {d(ax^{4}+b)}{dx}}}$

${\displaystyle =2x\;\;\;\;\;\;\;\;\;+4ax^{3}}$,

exactly as before.

If there were three functions of ${\displaystyle x}$, which we may call ${\displaystyle u}$, ${\displaystyle v}$ and ${\displaystyle w}$, so that

${\displaystyle y=u+v+w}$;

then

${\displaystyle {\frac {dy}{dx}}={\frac {du}{dx}}+{\frac {dv}{dx}}+{\frac {dw}{dx}}}$.

As for subtraction, it follows at once; for if the function ${\displaystyle v}$ had itself had a negative sign, its differential coefficient would also be negative; so that by differentiating

${\displaystyle y=u-v}$,

we should get

${\displaystyle {\frac {dy}{dx}}={\frac {du}{dx}}-{\frac {dv}{dx}}}$.

But when we come to do with Products, the thing is not quite so simple.

Suppose we were asked to differentiate the expression

${\displaystyle y=(x^{2}+c)\times (ax^{4}+b)}$,

what are we to do? The result will certainly not be ${\displaystyle 2x\times 4ax^{3}}$; for it is easy to see that neither ${\displaystyle c\times ax^{4}}$, nor ${\displaystyle x^{2}\times b}$, would have been taken into that product.

Now there are two ways in which we may go to work.

First way. Do the multiplying first, and, having worked it out, then differentiate.

Accordingly, we multiply together ${\displaystyle x^{2}+c}$ and ${\displaystyle ax^{4}+b}$.

This gives ${\displaystyle ax^{6}+acx^{4}+bx^{2}+bc}$.

Now differentiate, and we get:

${\displaystyle {\frac {dy}{dx}}=6ax^{5}+4acx^{3}+2bx}$.

Second way. Go back to first principles, and consider the equation

${\displaystyle y=u\times v}$;

where ${\displaystyle u}$ is one function of ${\displaystyle x}$, and ${\displaystyle v}$ is any other function of ${\displaystyle x}$. Then, if ${\displaystyle x}$ grows to be ${\displaystyle x+dx}$; and ${\displaystyle y}$ to ${\displaystyle y+dy}$; and ${\displaystyle u}$ becomes ${\displaystyle u+du}$, and ${\displaystyle v}$ becomes ${\displaystyle v+dv}$, we shall have:

${\displaystyle y+dy=(u+du)\times (v+dv)}$

${\displaystyle =u\cdot v+u\cdot dv+v\cdot du+du\cdot dv}$.

Now ${\displaystyle du\cdot dv}$ is a small quantity of the second order of smallness, and therefore in the limit may be discarded, leaving

${\displaystyle y+dy=u\cdot v+u\cdot dv+v\cdot du}$.

Then, subtracting the original ${\displaystyle y=u\cdot v}$, we have left

${\displaystyle dy=u\cdot dv+v\cdot du}$;

and, dividing through by ${\displaystyle dx}$, we get the result:

${\displaystyle {\frac {dy}{dx}}=u{\frac {dv}{dx}}+v{\frac {du}{dx}}}$.

This shows that our instructions will be as follows: 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.

You should note that this process amounts to the following: Treat ${\displaystyle u}$ as constant while you differentiate ${\displaystyle v}$; then treat ${\displaystyle v}$ as constant while you differentiate u; and the whole differential coefficient ${\displaystyle {\frac {dy}{dx}}}$ will be the sum of these two treatments.

Now, having found this rule, apply it to the concrete example which was considered above.

We want to differentiate the product

${\displaystyle (x^{2}+c)\times (ax^{4}+b)}$.

Call ${\displaystyle (x^{2}+c)=u}$; and ${\displaystyle (ax^{4}+b)=v}$.

Then, by the general rule just established, we may write:

${\displaystyle {\frac {dy}{dx}}=(x^{2}+c){\frac {d(ax^{4}+b)}{dx}}+(ax^{4}+b){\frac {d(x^{2}+c)}{dx}}}$

${\displaystyle =(x^{2}+c)4ax^{3}+(ax^{4}+b)2x}$

${\displaystyle =4ax^{5}+4acx^{3}+2ax^{5}+2bx}$,

${\displaystyle {\frac {dy}{dx}}=6ax^{5}+4acx^{3}+2bx}$,

exactly as before.

Lastly, we have to differentiate quotients.

Think of this example, ${\displaystyle y={\dfrac {bx^{5}+c}{x^{2}+a}}}$. In such a case it is no use to try to work out the division beforehand, because ${\displaystyle x^{2}+a}$ will not divide into ${\displaystyle bx^{5}+c}$, neither have they any common factor. So there is nothing for it but to go back to first principles, and find a rule.

So we will put ${\displaystyle y={\frac {u}{v}}}$;

where ${\displaystyle u}$ and ${\displaystyle v}$ are two different functions of the independent variable ${\displaystyle x}$. Then, when ${\displaystyle x}$ becomes ${\displaystyle x+dx}$, ${\displaystyle y}$ will become ${\displaystyle y+dy}$; and ${\displaystyle u}$ will become ${\displaystyle u+du}$; and ${\displaystyle v}$ will become ${\displaystyle v+dv}$. So then

${\displaystyle y+dy={\dfrac {u+du}{v+dv}}}$.

Now perform the algebraic division, thus:

${\displaystyle v+dv\,\,}$ ${\displaystyle u+du}$ ${\displaystyle {\dfrac {u}{v}}+{\dfrac {du}{v}}-{\dfrac {u\cdot dv}{v^{2}}}}$
${\displaystyle u+{\dfrac {u\cdot dv}{v}}}$
 ${\displaystyle du-{\dfrac {u\cdot dv}{v}}}$ ${\displaystyle du+{\dfrac {du\cdot dv}{v}}}$
 ⁠ ${\displaystyle -{\dfrac {u\cdot dv}{v}}-{\dfrac {du\cdot dv}{v}}}$ ${\displaystyle -{\dfrac {u\cdot dv}{v}}-{\dfrac {u\cdot dv\cdot dv}{v^{2}}}}$
 ⁠ ${\displaystyle -{\dfrac {du\cdot dv}{v}}+{\dfrac {u\cdot dv\cdot dv}{v^{2}}}}$.
As both these remainders are small quantities of the second order, they may be neglected, and the division may stop here, since any further remainders would be of still smaller magnitudes.

So we have got:

${\displaystyle y+dy={\frac {u}{v}}+{\frac {du}{v}}-{\frac {u\cdot dv}{v^{2}}}}$;

which may be written

${\displaystyle ={\frac {u}{v}}+{\frac {v\cdot du-u\cdot dv}{v^{2}}}}$.

Now subtract the original ${\displaystyle y={\frac {u}{v}}}$, and we have left:

${\displaystyle dy={\frac {v\cdot du-u\cdot dv}{v^{2}}}}$;

whence

${\displaystyle {\frac {dy}{dx}}={\frac {v{\frac {du}{dx}}-u{\frac {dv}{dx}}}{v^{2}}}}$.

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.

Going back to our example ${\displaystyle y={\frac {bx^{5}+c}{x^{2}+a}}}$,

write

${\displaystyle bx^{5}+c=u}$;

and

${\displaystyle x^{2}+a=v}$.

Then

${\displaystyle {\frac {dy}{dx}}={\frac {(x^{2}+a)\;{\frac {d(bx^{5}+c)}{dx}}-(bx^{5}+c)\;{\frac {d(x^{2}+a)}{dx}}}{(x^{2}+a)^{2}}}}$

${\displaystyle ={\frac {(x^{2}+a)(5bx^{4})-(bx^{5}+c)(2x)}{(x^{2}+a)^{2}}}}$,

${\displaystyle {\frac {dy}{dx}}={\frac {3bx^{6}+5abx^{4}-2cx}{(x^{2}+a)^{2}}}}$. (Answer.)

The working out of quotients is often tedious, but there is nothing difficult about it.

Some further examples fully worked out are given hereafter.

(1) Differentiate ${\displaystyle y={\dfrac {a}{b^{2}}}x^{3}-{\frac {a^{2}}{b}}x+{\frac {a^{2}}{b^{2}}}}$.

Being a constant, ${\displaystyle a^{2}b^{2}}$ vanishes, and we have

${\displaystyle {\frac {dy}{dx}}={\frac {a}{b^{2}}}\times 3\times x^{3-1}-{\frac {a^{2}}{b}}\times 1\times x^{1-1}}$.

But ${\displaystyle x^{1-1}=x^{0}=1}$; so we get:

${\displaystyle {\frac {dy}{dx}}={\frac {3a}{b^{2}}}x^{2}-{\frac {a^{2}}{b}}}$.

(2) Differentiate ${\displaystyle y=2a{\sqrt {bx^{3}}}-{\frac {3b{\sqrt[{3}]{a}}}{x}}-2{\sqrt {ab}}}$.

Putting ${\displaystyle x}$ in the index form, we get

${\displaystyle y=2a{\sqrt {b}}x^{\frac {3}{2}}-3b{\sqrt[{3}]{a}}x^{-1}-2{\sqrt {ab}}.}$.

Now

${\displaystyle {\frac {dy}{dx}}=2a{\sqrt {b}}\times {\tfrac {3}{2}}\times x^{{\frac {3}{2}}-1}-3b{\sqrt[{3}]{a}}\times (-1)\times x^{-1-1}}$;

or,

${\displaystyle {\frac {dy}{dx}}=3a{\sqrt {bx}}+{\frac {3b{\sqrt[{3}]{a}}}{x^{2}}}}$.

(3) Differentiate ${\displaystyle z=1.8{\sqrt[{3}]{\dfrac {1}{\theta ^{2}}}}-{\dfrac {4.4}{\sqrt[{5}]{\theta }}}-27^{\circ }}$.

This may be written: ${\displaystyle z=1.8\theta ^{-{\frac {2}{3}}}-4.4\theta ^{-{\frac {1}{5}}}-27^{\circ }}$

The ${\displaystyle 27^{\circ }}$ vanishes, and we have

${\displaystyle {\frac {dz}{d\theta }}=1.8\times -{\tfrac {2}{3}}\times \theta ^{-{\frac {2}{3}}-1}-4.4\times \left(-{\tfrac {1}{5}}\right)\theta ^{-{\frac {1}{5}}-1}}$;

or,

${\displaystyle {\frac {dz}{d\theta }}=-1.2\theta ^{-{\frac {5}{3}}}+0.88\theta ^{-{\frac {6}{5}}}}$;

or,

${\displaystyle {\frac {dz}{d\theta }}={\frac {0.88}{\sqrt[{5}]{\theta ^{6}}}}-{\frac {1.2}{\sqrt[{3}]{\theta ^{5}}}}}$.

(4) Differentiate ${\displaystyle v=(3t^{2}-1.2t+1)^{3}}$ A direct way of doing this will be explained later (see p.67); but we can nevertheless manage it now without any difficulty.

Developing the cube, we get

${\displaystyle v=27t^{6}-32.4t^{5}+39.96t^{4}-23.328t^{3}+13.32t^{2}-3.6t+1}$;

hence

${\displaystyle {\frac {dv}{dt}}=162t^{5}-162t^{4}+159.84t^{3}-69.984t^{2}+26.64t-3.6}$.

(5) Differentiate ${\displaystyle y=(2x-3)(x+1)^{2}}$

${\displaystyle {\frac {dy}{dx}}=(2x-3){\frac {d\left[(x+1)(x+1)\right]}{dx}}+(x+1)^{2}{\frac {d(2x-3)}{dx}}}$

${\displaystyle =(2x-3)\left[(x+1){\frac {d(x+1)}{dx}}\right.+\left.(x+1){\frac {d(x+1)}{dx}}\right]}$

${\displaystyle +(x+1)^{2}{\frac {d(2x-3)}{dx}}}$

${\displaystyle =2(x+1)\left[(2x-3)+(x+1)\right]=2(x+1)(3x-2)}$;

or, more simply, multiply out and then differentiate.

(6) Differentiate ${\displaystyle y=0.5x^{3}(x-3)}$

 ${\displaystyle {\frac {dy}{dx}}}$ ${\displaystyle =0.5\left[x^{3}{\frac {d(x-3)}{dx}}+(x-3){\frac {d(x^{3})}{dx}}\right]}$ ${\displaystyle =0.5\left[x^{3}+(x-3)\times 3x^{2}\right]=2x^{3}-4.5x^{2}}$.

Same remarks as for preceding example.

(7) Differentiate ${\displaystyle w=\left(\theta +{\dfrac {1}{\theta }}\right)\left({\sqrt {\theta }}+{\dfrac {1}{\sqrt {\theta }}}\right)}$

This may be written

${\displaystyle w=(\theta +\theta ^{-1})(\theta ^{\frac {1}{2}}+\theta ^{-{\frac {1}{2}}})}$.

 ${\displaystyle {\frac {dw}{d\theta }}}$ ${\displaystyle =(\theta +\theta ^{-1}){\frac {d(\theta ^{\frac {1}{2}}+\theta ^{-{\frac {1}{2}}})}{d\theta }}+(\theta ^{\frac {1}{2}}+\theta ^{-{\frac {1}{2}}}){\frac {d(\theta +\theta ^{-1})}{d\theta }}}$ ${\displaystyle =(\theta +\theta ^{-1})({\tfrac {1}{2}}\theta ^{-{\frac {1}{2}}}-{\tfrac {1}{2}}\theta ^{-{\frac {3}{2}}})+(\theta ^{\frac {1}{2}}+\theta ^{-{\frac {1}{2}}})(1-\theta ^{-2})}$ ${\displaystyle ={\tfrac {1}{2}}(\theta ^{\frac {1}{2}}+\theta ^{-{\frac {3}{2}}}-\theta ^{-{\frac {1}{2}}}-\theta ^{-{\frac {5}{2}}})+(\theta ^{\frac {1}{2}}+\theta ^{-{\frac {1}{2}}}-\theta ^{-{\frac {3}{2}}}-\theta ^{-{\frac {5}{2}}})}$ ${\displaystyle ={\tfrac {3}{2}}\left({\sqrt {\theta }}-{\frac {1}{\sqrt {\theta ^{5}}}}\right)+{\tfrac {1}{2}}\left({\frac {1}{\sqrt {\theta }}}-{\frac {1}{\sqrt {\theta ^{3}}}}\right)}$.

This, again, could be obtained more simply by multiplying the two factors first, and differentiating afterwards. This is not, however, always possible; see, for instance, p. 173, example 8, in which the rule for differentiating a product must be used.

(8) Differentiate ${\displaystyle y={\frac {a}{1+a{\sqrt {x}}+a^{2}x}}}$.

 ${\displaystyle {\frac {dy}{dx}}}$ ${\displaystyle ={\frac {(1+ax^{\frac {1}{2}}+a^{2}x)\times 0-a{\frac {d(1+ax^{\frac {1}{2}}+a^{2}x)}{dx}}}{(1+a{\sqrt {x}}+a^{2}x)^{2}}}}$ ${\displaystyle =-{\frac {a({\frac {1}{2}}ax^{-{\frac {1}{2}}}+a^{2})}{(1+ax^{\frac {1}{2}}+a^{2}x)^{2}}}}$.

(9) Differentiate ${\displaystyle y={\frac {x^{2}}{x^{2}+1}}}$.

${\displaystyle {\frac {dy}{dx}}={\frac {(x^{2}+1)\,2x-x^{2}\times 2x}{(x^{2}+1)^{2}}}={\frac {2x}{(x^{2}+1)^{2}}}}$.

(10) Differentiate

${\displaystyle y={\frac {a+{\sqrt {x}}}{a-{\sqrt {x}}}}}$.

In the indexed form,

${\displaystyle y={\frac {a+x^{\frac {1}{2}}}{a-x^{\frac {1}{2}}}}}$.

${\displaystyle {\frac {dy}{dx}}={\frac {(a-x^{\frac {1}{2}})({\tfrac {1}{2}}x^{-{\frac {1}{2}}})-(a+x^{\frac {1}{2}})(-{\tfrac {1}{2}}x^{-{\frac {1}{2}}})}{(a-x^{\frac {1}{2}})^{2}}}={\frac {a-x^{\frac {1}{2}}+a+x^{\frac {1}{2}}}{2(a-x^{\frac {1}{2}})^{2}x^{\frac {1}{2}}}}}$;

hence

${\displaystyle {\frac {dy}{dx}}={\frac {a}{(a-{\sqrt {x}})^{2}\,{\sqrt {x}}}}}$.

(11) Differentiate

${\displaystyle \theta ={\frac {1-a{\sqrt[{3}]{t^{2}}}}{1+a{\sqrt[{2}]{t^{3}}}}}}$.

Now

${\displaystyle \theta ={\frac {1-at^{\frac {2}{3}}}{1+at^{\frac {3}{2}}}}}$.

 ${\displaystyle {\frac {d\theta }{dt}}}$ ${\displaystyle ={\frac {(1+at^{\frac {3}{2}})(-{\tfrac {2}{3}}at^{-{\frac {1}{3}}})-(1-at^{\frac {2}{3}})\times {\tfrac {3}{2}}at^{\frac {1}{2}}}{(1+at^{\frac {3}{2}})^{2}}}}$ ${\displaystyle ={\frac {5a^{2}{\sqrt[{6}]{t^{7}}}-{\dfrac {4a}{\sqrt[{3}]{t}}}-9a{\sqrt[{2}]{t}}}{6(1+a{\sqrt[{2}]{t^{3}}})^{2}}}}$.

(12) A reservoir of square cross-section has sides sloping at an angle of ${\displaystyle 45^{\circ }}$ with the vertical. The side of the bottom is ${\displaystyle 200}$ feet. Find an expression for the quantity pouring in or out when the depth of water varies by ${\displaystyle 1}$ foot; hence find, in gallons, the quantity withdrawn hourly when the depth is reduced from ${\displaystyle 14}$ to ${\displaystyle 10}$ feet in ${\displaystyle 24}$ hours.

The volume of a frustum of pyramid of height ${\displaystyle H}$, and of bases ${\displaystyle A}$ and ${\displaystyle a}$, is ${\displaystyle V={\dfrac {H}{3}}(A+a+{\sqrt {Aa}})}$. It is easily seen that, the slope being ${\displaystyle 45^{\circ }}$, if the depth be ${\displaystyle h}$, the length of the side of the square surface of the water is ${\displaystyle 200+2h}$ feet, so that the volume of water is

${\displaystyle {\dfrac {h}{3}}[200^{2}+(200+2h)^{2}+200(200+2h)]=40,000h+400h^{2}+{\dfrac {4h^{3}}{3}}}$.

${\displaystyle {\dfrac {dV}{dh}}=40,000+800h+4h^{2}=}$ cubic feet per foot of depth variation. The mean level from ${\displaystyle 14}$ to ${\displaystyle 10}$ feet is ${\displaystyle 12}$ feet, when ${\displaystyle h=12}$, ${\displaystyle {\dfrac {dV}{dh}}=50,176}$ cubic feet.

Gallons per hour corresponding to a change of depth of ${\displaystyle 4}$ ft. in ${\displaystyle 24}$ hours ${\displaystyle {}={\dfrac {4\times 50,176\times 6.25}{24}}=52,267}$ gallons.

(13) The absolute pressure, in atmospheres, ${\displaystyle P}$, of saturated steam at the temperature ${\displaystyle t^{\circ }}$ C. is given by Dulong as being ${\displaystyle P=\left({\dfrac {40+t}{140}}\right)^{5}}$ as long as ${\displaystyle t}$ is above 80°. Find the rate of variation of the pressure with the temperature at ${\displaystyle 100^{\circ }}$ C.

Expand the numerator by the binomial theorem (see p. 141).

${\displaystyle P={\frac {1}{140^{5}}}(40^{5}+5\times 40^{4}t+10\times 40^{3}t^{2}+10\times 40^{2}t^{3}+5\times 40t^{4}+t^{5})}$;

hence ${\displaystyle {\dfrac {dP}{dt}}={\dfrac {1}{537,824\times 10^{5}}}}$,

${\displaystyle (5\times 40^{4}+20\times 40^{3}t+30\times 40^{2}t^{2}+20\times 40t^{3}+5t^{4})}$,

when ${\displaystyle t=100}$ this becomes ${\displaystyle 0.036}$ atmosphere per degree Centigrade change of temperature.

Exercises III. (See the Answers on p. 255.)

(1) Differentiate

(a) ${\displaystyle u=1+x+{\dfrac {x^{2}}{1\times 2}}+{\dfrac {x^{3}}{1\times 2\times 3}}+\dotsb }$
(b) ${\displaystyle y=ax^{2}+bx+c}$
(c) ${\displaystyle y=(x+a)^{2}}$
(d) ${\displaystyle y=(x+a)^{3}}$

(2) If ${\displaystyle w=at-{\frac {1}{2}}bt^{2}}$, find ${\displaystyle {\dfrac {dw}{dt}}}$.

(3) Find the differential coefficient of

${\displaystyle y=(x+{\sqrt {-1}})\times (x-{\sqrt {-1}})}$.

(4) Differentiate

${\displaystyle y=(197x-34x^{2})\times (7+22x-83x^{3}).}$.

(5) If ${\displaystyle x=(y+3)\times (y+5)}$, find ${\displaystyle {\dfrac {dx}{dy}}}$.

(6) Differentiate ${\displaystyle y=1.3709x\times (112.6+45.202x^{2})}$

Find the differential coefficients of

 (7) ${\displaystyle y={\dfrac {2x+3}{3x+2}}}$. (8) ${\displaystyle y={\dfrac {1+x+2x^{2}+3x^{3}}{1+x+2x^{2}}}}$. (9) ${\displaystyle y={\dfrac {ax+b}{cx+d}}}$. (10) ${\displaystyle y={\dfrac {x^{n}+a}{x^{-n}+b}}}$.

(11) The temperature ${\displaystyle t}$ of the filament of an incandescent electric lamp is connected to the current passing through the lamp by the relation

${\displaystyle C=a+bt+ct^{2}}$.

Find an expression giving the variation of the current corresponding to a variation of temperature.

(12) The following formulae have been proposed to express the relation between the electric resistance ${\displaystyle R}$ of a wire at the temperature ${\displaystyle t^{\circ }}$ C., and the resistance ${\displaystyle R_{0}}$ of that same wire at ${\displaystyle 0^{\circ }}$ Centigrade, ${\displaystyle a}$, ${\displaystyle b}$, ${\displaystyle c}$ being constants.

${\displaystyle R=R_{0}(1+at+bt^{2})}$.
${\displaystyle R=R_{0}(1+at+b{\sqrt {t}})}$.
${\displaystyle R=R_{0}(1+at+bt^{2})^{-1}}$.

Find the rate of variation of the resistance with regard to temperature as given by each of these formulae.

(13) The electromotive-force ${\displaystyle E}$ of a certain type of standard cell has been found to vary with the temperature t according to the relation

${\displaystyle E=1.4340{\bigl [}1-0.000814(t-15)+0.000007(t-15)^{2}{\bigr ]}}$ volts.

Find the change of electromotive-force per degree, at ${\displaystyle 15^{\circ }}$, ${\displaystyle 20^{\circ }}$ and ${\displaystyle 25^{\circ }}$.

(14) The electromotive-force necessary to maintain an electric arc of length ${\displaystyle l}$ with a current of intensity ${\displaystyle i}$ has been found by Mrs. Ayrton to be

${\displaystyle E=a+bl+{\frac {c+kl}{i}}}$,

where ${\displaystyle a}$, ${\displaystyle b}$, ${\displaystyle c}$, ${\displaystyle k}$ are constants.

Find an expression for the variation of the electromotive-force ${\displaystyle (a)}$ with regard to the length of the arc; ${\displaystyle (b)}$ with regard to the strength of the current.