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: \[\begin{aligned}\dfrac{dy}{dx} &= \dfrac{du}{dx} + \dfrac{dv}{dx}.\end{aligned}\]

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 \[\begin{aligned} y &= u + v + w\\ \dfrac{dy}{dx} &= \dfrac{du}{dx} + \dfrac{dv}{dx} + \dfrac{dw}{dx},\end{aligned}\]

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 \[\begin{aligned} y &= u – v\\ \dfrac{dy}{dx} &= \dfrac{du}{dx} – \dfrac{dv}{dx},\end{aligned}\]

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 \[y = u \times v;\] 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: \[\dfrac{dy}{dx} = u\, \dfrac{dv}{dx} + v\, \dfrac{du}{dx}.\]

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 \[y = \frac{u}{v}; \]

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:\[\begin{aligned} dy &= \frac{v\cdot du-u\cdot dv}{v^2}; \\ \dfrac{dy}{dx} &= \frac{v\cdot \dfrac{du}{dx}-u\cdot \dfrac{dv}{dx}}{v^2}. \end{aligned}\]

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.