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:

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

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

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: \[ \dfrac{dy}{dx} = 6ax^5 + 4acx^3 + 2bx. \]

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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