Sums, Differences, Products and Quotients

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.

Subtracting the original $y = u+v$, we get

and dividing through by $dx$, we get:

$\dfrac{dy}{dx} = \dfrac{du}{dx} + \dfrac{dv}{dx}.$

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

If there were three functions of $x$, which we may call $u$, $v$ and $w$, so that \begin{align*} y &= u+v+w; \\ \text{then}\; \frac{dy}{dx} &= \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx}. \end{align*}

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
\begin{align*}
y &= u-v, \\
\text{ we should get}\;
\frac{dy}{dx} &= \frac{du}{dx} - \frac{dv}{dx}.
\end{align*}

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

Suppose we were asked to differentiate the expression
\[
y = (x^2+c) × (ax^4+b),
\]
what are we to do? 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.

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
\[
y = u × v;
\]
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{align*}
y + dy &= (u + du) × (v + dv) \\
&= u · v + u · dv + v · du + du · dv.
\end{align*}

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

Then, subtracting the original $y = u· v$, we have left \[ dy = u · dv + v · du; \] and, dividing through by $dx$, we get the result: \[ \dfrac{dy}{dx} = u\, \dfrac{dv}{dx} + v\, \dfrac{du}{dx}. \]

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.*

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

We want to differentiate the product \[ (x^2 + c) × (ax^4 + b). \]

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

Lastly, we have to differentiate *quotients*.

Now perform the algebraic division, thus:

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 $y = \dfrac{a}{b^2} x^3 - \dfrac{a^2}{b} x + \dfrac{a^2}{b^2}$.

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

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

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

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

(4) Differentiate $v = (3t^2 - 1.2 t + 1)^3$.

A direct way of doing this will be explained later (see here); 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 \[ \frac{dv}{dt} = 162t^5 - 162t^4 + 159.84t^3 - 69.984t^2 + 26.64t - 3.6. \]

(5) Differentiate $y = (2x - 3)(x + 1)^2$. \begin{alignat*}{2} \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} \\ &= 2(x + 1)\bigl[(2x - 3) + (x + 1)\bigr] &&= 2(x + 1)(3x - 2) \end{alignat*} or, more simply, multiply out and then differentiate.

(6) Differentiate $y = 0.5 x^3(x-3)$. \begin{align*} \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) × 3x^2\right] = 2x^3 - 4.5x^2. \end{align*}

Same remarks as for preceding example.

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

This may be written \begin{gather*} w = (\theta + \theta^{-1})(\theta^{\frac{1}{2}} + \theta^{-\frac{1}{2}}). \\ \begin{aligned} \frac{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} \end{gather*}

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

(8) Differentiate $y =\dfrac{a}{1 + a\sqrt{x} + a^2x}$. \begin{align*} \frac{dy}{dx} &= \frac{(1 + ax^{\frac{1}{2}} + a^2x) × 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{align*}

(9) Differentiate $y = \dfrac{x^2}{x^2 + 1}$. \[ \dfrac{dy}{dx} = \dfrac{(x^2 + 1)\, 2x - x^2 × 2x}{(x^2 + 1)^2} = \dfrac{2x}{(x^2 + 1)^2}. \]

(10) Differentiate $y = \dfrac{a + \sqrt{x}}{a - \sqrt{x}}$.

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

(11) Differentiate

\begin{align*} \theta &= \frac{1 - a \sqrt[3]{t^2}}{1 + a \sqrt[2]{t^3}}. \\ \text{Now}\; \theta &= \frac{1 - at^{\frac{2}{3}}}{1 + at^{\frac{3}{2}}}. \end{align*} \begin{align*} \frac{d\theta}{dt} &= \frac{(1 + at^{\frac{3}{2}}) (-\tfrac{2}{3} at^{-\frac{1}{3}}) - (1 - at^{\frac{2}{3}}) × \tfrac{3}{2} at^{\frac{1}{2}}} {(1 + at^{\frac{3}{2}})^2} \\ &= \frac{5a^2 \sqrt[6]{t^7} - \dfrac{4a}{\sqrt[3]{t}} - 9a \sqrt[2]{t}} {6(1 + a \sqrt[2]{t^3})^2}. \end{align*}

(12) A reservoir of square cross-section has sides sloping at an angle of $45°$ 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.

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 $45°$, 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 ${} = \dfrac{4 × 50,176 × 6.25}{24} = 52,267$ gallons.

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

Expand the numerator by the binomial theorem (see here). \[ P = \frac{1}{140^5} (40^5 + 5×40^4 t + 10 × 40^3 t^2 + 10 × 40^2 t^3 + 5 × 40t^4 + t^5); \] \begin{align*} \text{hence}\; \dfrac{dP}{dt} = &\dfrac{1}{537,824 × 10^5}\\ &(5 × 40^4 + 20 × 40^3 t + 30 × 40^2 t^2 + 20 × 40t^3 + 5t^4), \end{align*} when $t = 100$ this becomes $0.036$ atmosphere per degree Centigrade change of temperature.

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

(*b*) $y = ax^2 + bx + c$. (*c* ) $y = (x + a)^2$.

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

(3) Find the differential coefficient of \[ y = (x + \sqrt{-1}) × (x - \sqrt{-1}). \]

(4) Differentiate \[ y = (197x - 34x^2) × (7 + 22x - 83x^3). \]

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

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

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) $y = \dfrac{x^n + a}{x^{-n} + b}$.

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

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

(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) $1.4340(0.000014t - 0.001024)$, $-0.00117$, $-0.00107$, $-0.00097$.

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

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