Page:Calculus Made Easy.pdf/39

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Simplest Cases

19

What does $(dx)^{2}$ mean? Remember that $dx$ meant a bit–a little bit–of $x$ . Then $(dx)^{2}$ will mean a little bit of a little bit of $x$ ; that is, as explained above (p. 4), it is a small quantity of the second order of smallness. It may therefore be discarded as quite inconsiderable in comparison with the other terms. Leaving it out, we then have:

$y+dy=x^{2}+2x\cdot dx$ .

Now $y=x^{2}$ ; so let us subtract this from the equation and we have left

$dy=2x\cdot dx$ .

Dividing across by $dx$ , we find

${\frac {dy}{dx}}=2x$ .

Now this^[1] is what we set out to find. The ratio of the growing of $y$ to the growing of $x$ is, in the case before us, found to be $2x$ .

↑ N.B.—This ratio ${\frac {dy}{dx}}$ is the result of differentiating $y$ with respect to $x$ . Differentiating means finding the differential coefficient. Suppose we had some other function of $x$ , as, for example, $u=7x^{2}+3$ . Then if we were told to differentiate this with respect to $x$ , we should have to find ${\frac {du}{dx}}$ , or, what is the same thing, ${\frac {d(7x^{2}+3)}{dx}}$ . On the other hand, we may have a case in which time was the independent variable (see p. 15), such as this: $y=b+{\tfrac {1}{2}}at^{2}$ . Then, if we were told to differentiate it, that means we must find its differential coefficient with respect to $t$ . So that then our business would be to try to find ${\frac {dy}{dt}}$ , that is, to find ${\frac {d(b+{\tfrac {1}{2}}at^{2})}{dt}}$ .

[1] N.B.—This ratio ${\frac {dy}{dx}}$ is the result of differentiating $y$ with respect to $x$ . Differentiating means finding the differential coefficient. Suppose we had some other function of $x$ , as, for example, $u=7x^{2}+3$ . Then if we were told to differentiate this with respect to $x$ , we should have to find ${\frac {du}{dx}}$ , or, what is the same thing, ${\frac {d(7x^{2}+3)}{dx}}$ . On the other hand, we may have a case in which time was the independent variable (see p. 15), such as this: $y=b+{\tfrac {1}{2}}at^{2}$ . Then, if we were told to differentiate it, that means we must find its differential coefficient with respect to $t$ . So that then our business would be to try to find ${\frac {dy}{dt}}$ , that is, to find ${\frac {d(b+{\tfrac {1}{2}}at^{2})}{dt}}$ .

[1]