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

For instance, let $y = (x^2+c) + (ax^4+b);$ what will its $$\dfrac{dy}{dx}$$ 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: $\dfrac{dy}{dx} = 2x + 4ax^3.\quad (\textit{Ans}.)$

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

And we shall have: \begin{aligned} y+dy &= u+du + v+dv.\end{aligned}

Subtracting the original $$y = u+v$$, we get \begin{aligned}dy &= du+dv,\end{aligned}

and dividing through by $$dx$$, we get: \bbox[#F2F2F2,5px,border:2px solid black]{\begin{aligned}\dfrac{dy}{dx} &= \dfrac{du}{dx} + \dfrac{dv}{dx}.\end{aligned}}

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 (see chapter 3), \begin{aligned} {2} \frac{dy}{dx} & = \frac{d(x^2+c)}{dx} &&+ \frac{d(ax^4+b)}{dx} \\ & = 2x &&+ 4ax^3,\end{aligned} exactly as before.

If there were three functions of $$x$$, which we may call $$u$$, $$v$$ and $$w$$, so that \bbox[#F2F2F2,5px,border:2px solid black]{\begin{aligned} y &= u + v + w\\ \dfrac{dy}{dx} &= \dfrac{du}{dx} + \dfrac{dv}{dx} + \dfrac{dw}{dx},\end{aligned}}

As for subtraction, it follows at once; for if the function $$v$$ had itself had a negative sign, its differential coefficient would also be negative; so that by differentiating \bbox[#F2F2F2,5px,border:2px solid black]{\begin{aligned} y &= u – v\\ \dfrac{dy}{dx} &= \dfrac{du}{dx} – \dfrac{dv}{dx},\end{aligned}}

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

Suppose we were asked to differentiate the expression $y = (x^2+c) \times (ax^4+b),$ what are we to do? The result will certainly not be $$2x \times 4ax^3$$; for it is easy to see that neither $$c \times ax^4$$, nor $$x^2 \times b$$, 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 $$x^2 + c$$ and $$ax^4 + b$$.

This gives $$ax^6 + acx^4 + bx^2 + bc$$.

Now differentiate, and we get: $\dfrac{dy}{dx} = 6ax^5 + 4acx^3 + 2bx.$

Second way, Go back to first principles, and consider the equation $\bbox[#F2F2F2,5px,border:2px solid black]{y = u \times v;}$ where $$u$$ is one function of $$x$$, and $$v$$ is any other function of $$x$$. Then, if $$x$$ grows to be $$x+dx$$; and $$y$$ to $$y+dy$$; and $$u$$ becomes $$u+du$$, and $$v$$ becomes $$v+dv$$, we shall have: \begin{aligned} y + dy &= (u + du) \times (v + dv) \\ &= u \cdot v + u \cdot dv + v \cdot du + du \cdot dv.\end{aligned}

Now $$du \cdot dv$$ is a small quantity of the second order of smallness, and therefore in the limit may be discarded, leaving $y + dy = u \cdot v + u \cdot dv + v \cdot du.$

Then, subtracting the original $$y = \cdot v$$, we have left $dy = u \cdot dv + v \cdot du;$ and, dividing through by $$dx$$, we get the result: $\bbox[#F2F2F2,5px,border:2px solid black]{\dfrac{dy}{dx} = u\, \dfrac{dv}{dx} + v\, \dfrac{du}{dx}.}$

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 $$u$$ 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.

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

We want to differentiate the product $(x^2 + c) \times (ax^4 + b).$

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

Then, by the general rule just established, we may write: \begin{aligned} \dfrac{dy}{dx} &= (x^2 + c)\, \frac{d(ax^4 + b)}{dx} &&+ (ax^4 + b)\, \frac{d(x^2 + c)}{dx} \\ &= (x^2 + c)\, 4ax^3 &&+ (ax^4 + b)\, 2x \\ &= 4ax^5 + 4acx^3 &&+ 2ax^5 + 2bx, \\ \dfrac{dy}{dx} &= 6ax^5 + 4acx^3 &&+ 2bx,\end{aligned} exactly as before.

Lastly, we have to differentiate quotients.

Think of this example, $$y = \dfrac{bx^5 + c}{x^2 + a}$$. In such a case it is no use to try to work out the division beforehand, because $$x^2 + a$$ will not divide into $$bx^5 + c$$, 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 $\bbox[#F2F2F2,5px,border:2px solid black]{y = \frac{u}{v};}$

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

Now perform the algebraic division, thus: 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:\begin{aligned} y+dy &= \frac{u}{v} + \frac{du}{v} – \frac{u\cdot dv}{v^2}; \\ &= \frac{u}{v} + \frac{v\cdot du-u\cdot dv}{v^2}. \end{aligned}

Now subtract the original $$y =\dfrac{u}{v}$$, and we have left:$dy = \frac{v\cdot du-u\cdot dv}{v^2};$\bbox[#F2F2F2,5px,border:2px solid black]{\begin{aligned} \dfrac{dy}{dx} = \frac{v\cdot \dfrac{du}{dx}-u\cdot \dfrac{dv}{dx}}{v^2}. \end{aligned}}

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 $$y = \dfrac{bx^5 + c}{x^2 + a}$$,

Write                                                         $bx^5 + c = u;$ and                                                              $x^2 + a = v.$

Then \begin{aligned} \frac{dy}{dx} &= \frac{(x^2 + a)\, \dfrac{d(bx^5 + c)}{dx} – (bx^5 + c)\, \dfrac{d(x^2 + a)}{dx}}{(x^2 + a)^2} \\ &= \frac{(x^2 + a)(5bx^4) – (bx^5 + c)(2x)}{(x^2 + a)^2}, \\ \frac{dy}{dx} &= \frac{3bx^6 + 5abx^4 – 2cx}{(x^2 + a)^2}.\end{aligned}

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

Some further examples fully worked out are given hereafter.

Example 1
Differentiate $$y = \dfrac{a}{b^2} x^3 – \dfrac{a^2}{b} x + \dfrac{a^2}{b^2}$$.
Solution

Being a constant, $$\dfrac{a^2}{b^2}$$ vanishes, and we have $\frac{dy}{dx} = \frac{a}{b^2} \times 3 \times x^{3-1} – \frac{a^2}{b} \times 1 \times x^{1-1}.$

But $$x^{1-1} = x^0 = 1$$; so we get: $\frac{dy}{dx} = \frac{3a}{b^2} x^2 – \frac{a^2}{b}.$

Example 2
Differentiate $$y = 2a\sqrt{bx^3} – \dfrac{3b \sqrt{a}}{x} – 2\sqrt{ab}$$.
Solution

Putting $$x$$ in the index form, we get $y = 2a\sqrt{b} x^{\frac{3}{2}} – 3b \sqrt{a} x^{-1} – 2\sqrt{ab}.$

Now \begin {aligned}\dfrac{dy}{dx} &= 2a\sqrt{b} \times \tfrac{3}{2} \times x^{\frac{3}{2}-1} – 3b\sqrt{a} \times (-1) \times x^{-1-1}; \\ \frac{dy}{dx} &= 3a\sqrt{bx} + \frac{3b\sqrt{a}}{x^2}.\end{aligned}

Example 3
Differentiate $$z = 1.8 \sqrt{\dfrac{1}{\theta^2}} – \dfrac{4.4}{\sqrt{\theta}} – 27^\circ$$.
Solution

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

The $$27^\circ$$ vanishes, and we have \begin{aligned} \dfrac{dz}{d\theta} &= 1.8 \times -\frac{2}{3} \times \theta^{-\frac{2}{3}-1} – 4.4 \times \left(-\frac{1}{5}\right)\theta^{-\frac{1}{5}-1}; \\ \dfrac{dz}{d\theta} &= -1.2\, \theta^{-\frac{5}{3}} + 0.88\, \theta^{-\frac{6}{5}}; \\ \dfrac{dz}{d\theta} &= \frac{0.88}{\sqrt{\theta^6}} – \frac{1.2}{\sqrt{\theta^5}}. \end{aligned}

Example 4
Differentiate $$v = (3t^2 – 1.2 t + 1)^3$$.
Solution

A direct way of doing this will be explained later (see chapter 9); but we can nevertheless manage it now without any difficulty.

Developing the cube, we get $v = 27t^6 – 32.4t^5 + 39.96t^4 – 23.328t^3 – 13.32t^2 – 3.6t + 1;$ hence $\frac{dv}{dt} = 162t^5 – 162t^4 + 159.84t^3 – 69.984t^2 + 26.64t – 3.6.$

Example 5
Differentiate $$y = (2x – 3)(x + 1)^2$$.
Solution

\begin{aligned} \frac{dy}{dx} &= (2x – 3)\, \frac{d\bigl[(x + 1)(x + 1)\bigr]}{dx} &&+ (x + 1)^2\, \frac{d(2x – 3)}{dx} \\ &= (2x – 3) \left[(x + 1)\, \frac{d(x + 1)}{dx}\right. &&+ \left.(x + 1)\, \frac{d(x + 1)}{dx}\right] \\ & &&+ (x + 1)^2\, \frac{d(2x – 3)}{dx} \\ &= \rlap{2(x + 1)\bigl[(2x – 3) + (x + 1)\bigr] = 2(x + 1)(3x – 2);}&&\end{aligned}
or, more simply, multiply out and then differentiate.

Example 6
Differentiate $$y = 0.5 x^3(x-3)$$.
Solution

\begin{aligned} \frac{dy}{dx} &= 0.5\left[x^3 \frac{d(x-3)}{dx} + (x-3) \frac{d(x^3)}{dx}\right] \\ &= 0.5\left[x^3 + (x-3) \times 3x^2\right] = 2x^3 – 4.5x^2.\end{aligned}

Same remarks as for preceding example.

Example 7
Differentiate $$w = \left(\theta + \dfrac{1}{\theta}\right) \left(\sqrt{\theta} + \dfrac{1}{\sqrt{\theta}}\right)$$.
Solution
This may be written $w = (\theta + \theta^{-1})(\theta^{\frac{1}{2}} + \theta^{-\frac{1}{2}}).$

\begin{aligned} \dfrac{dw}{d\theta} &= (\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} \\ &= (\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}) \\ &= \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}}) \\ &= \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).\end{aligned}

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, , example 8, in which the rule for differentiating a product must be used.

Example 8
Differentiate $$y =\dfrac{a}{1 + a\sqrt{x} + a^2x}$$.
Solution

\begin{aligned} \frac{dy}{dx} &= \frac{(1 + ax^{\frac{1}{2}} + a^2x) \times 0 – a\dfrac{d(1 + ax^{\frac{1}{2}} + a^2x)}{dx}} {(1 + a\sqrt{x} + a^2x)^2} \\ &= – \frac{a(\frac{1}{2}ax^{-\frac{1}{2}} + a^2)} {(1 + ax^{\frac{1}{2}} + a^2x)^2}.\end{aligned}

Example 9
Differentiate $$y = \dfrac{x^2}{x^2 + 1}$$.
Solution

$\dfrac{dy}{dx} = \dfrac{(x^2 + 1)\, 2x – x^2 \times 2x}{(x^2 + 1)^2} = \dfrac{2x}{(x^2 + 1)^2}.$

Example 10
Differentiate $$y = \dfrac{a + \sqrt{x}}{a – \sqrt{x}}$$.
Solution

In the indexed form, $$y = \dfrac{a + x^{\frac{1}{2}}}{a – x^{\frac{1}{2}}}$$. $\frac{dy}{dx} = \frac{(a – x^{\frac{1}{2}})( \frac{1}{2} x^{-\frac{1}{2}}) – (a + x^{\frac{1}{2}})(-\tfrac{1}{2} x^{-\frac{1}{2}})}{(a – x^{\frac{1}{2}})^2} = \frac{ a – x^{\frac{1}{2}} + a + x^{\frac{1}{2}}}{2(a – x^{\frac{1}{2}})^2\, x^{\frac{1}{2}}};$

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

Example 11
Differentiate $$\theta = \dfrac{1 – a \sqrt{t^2}}{1 + a \sqrt{t^3}}= \dfrac{1 – at^{\frac{2}{3}}}{1 + at^{\frac{3}{2}}}.$$

Solution

\begin{aligned} \frac{d\theta}{dt} &= \frac{(1 + at^{\frac{3}{2}}) (-\frac{2}{3} at^{-\frac{1}{3}}) – (1 – at^{\frac{2}{3}}) \times \frac{3}{2} at^{\frac{1}{2}}} {(1 + at^{\frac{3}{2}})^2} \\ &= \frac{5a^2 \sqrt{t^7} – \dfrac{4a}{\sqrt{t}} – 9a \sqrt{t}} {6(1 + a \sqrt{{3}{t^3}})^2}.\end{aligned}

Example 12

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

Solution

The volume of a frustum of pyramid of height $$H$$, and of bases $$A$$ and $$a$$, is $$V = \dfrac{H}{3} (A + a + \sqrt{Aa} )$$. It is easily seen that, the slope being $$45^\circ$$, if the depth be $$h$$, the length of the side of the square surface of the water is $$200 + 2h$$ feet, so that the volume of water is $\dfrac{h}{3} [200^2 + (200 + 2h)^2 + 200(200 + 2h)] = 40,000h + 400h^2 + \dfrac{4h^3}{3}.$

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

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

Example 13

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

Solution

Expand the numerator by the binomial theorem (see chapter 14). $P = \frac{1}{140^5} (40^5 + 5\times 40^4 t + 10 \times 40^3 t^2 + 10 \times 40^2 t^3 + 5 \times 40t^4 + t^5);$

hence \begin{aligned} \dfrac{dP}{dt} = & \frac{1}{537.824 \times 10^5} \\ &(5 ^4 + 20 ^3 t + 30 ^2 t^2 + 20 40t^3 + 5t^4),\end{aligned}

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

Exercises III.

(1) Differentiate

(a) $$u = 1 + x + \dfrac{x^2}{1 \times 2} + \dfrac{x^3}{1 \times 2 \times 3} + \dotsb$$.

 (b) $$y = ax^2 + bx + c$$ (c) $$y = (x + a)^2$$ (d) $$y = (x + a)^3$$

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

(3) Find the differential coefficient of $y = (x + \sqrt{-1}) \times (x – \sqrt{-1}).$

(4) Differentiate $y = (197x – 34x^2) \times (7 + 22x – 83x^3).$

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

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

Find the differential coefficients of

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

(11) 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.

(12) The following formulae have been proposed to express the relation between the electric resistance $$R$$ of a wire at the temperature $$t^\circ$$C., and the resistance $$R_0$$ of that same wire at $$0^\circ$$ Centigrade, $$a$$, $$b$$, $$c$$ being constants. \begin{aligned} R &= R_0(1 + at + bt^2). \\ R &= R_0(1 + at + b\sqrt{t}). \\ R &= R_0(1 + at + bt^2)^{-1}.\end{aligned}

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

(13) The electromotive-force $$E$$ of a certain type of standard cell has been found to vary with the temperature $$t$$ according to the relation $E = 1.4340 \bigl[1 – 0.000814(t-15) + 0.000007(t-15)^2\bigr] \text{ volts}.$

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

(14) 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 $E = a + bl + \frac{c + kl}{i},$ 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.[/su_spoiler]

(1)

 (a) $$1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \dfrac{x^4}{24} + \ldots$$ (b) $$2ax + b$$ (c) $$2x + 2a$$ (d) $$3x^2 + 6ax + 3a^2$$
 (2) $$\dfrac{dw}{dt} = a – bt$$ (3) $$\dfrac{dy}{dx} = 2x$$

(4) $$14110x^4 – 65404x^3 – 2244x^2 + 8192x + 1379$$.

 (5) $$\dfrac{dx}{dy} = 2y + 8$$ (6) $$185.9022654x^2 + 154.36334$$ (7) $$\dfrac{-5}{(3x + 2)^2}$$ (8) $$\dfrac{6x^4 + 6x^3 + 9x^2}{(1 + x + 2x^2)^2}$$ (9) $$\dfrac{ad – bc}{(cx + d)^2}$$ (10) $$\dfrac{anx^{-n-1} + bnx^{n-1} + 2nx^{-1}}{(x^{-n} + b)^2}$$

(11) $$b + 2ct$$.

(12) $$R_0(a + 2bt)$$,$$R_0 \left(a + \dfrac{b}{2\sqrt{t}}\right)$$, $$-\dfrac{R_0(a + 2bt)}{(1 + at + bt^2)^2}$$or$$\dfrac{R^2 (a + 2bt)}{R_0}$$.

(13) $$1.4340(0.000014t – {0.000828}{0.001024})$$,$$-0.00117$$,$$-0.00107$$,$$-0.00097$$.

(14) $$\dfrac{dE}{dl} = b + \dfrac{k}{i}$$,                           $$\dfrac{dE}{di} = -\dfrac{c + kl}{i^2}$$.