Page:Calculus Made Easy.pdf/40

Numerical example.

Suppose $x=100$ and ∴ $y=10,000$. Then let $x$ grow till it becomes $101$ (that is, let $dx=1$). Then the enlarged $y$ will be $101\times 101=10,201$. But if we agree that we may ignore small quantities of the second order, $1$ may be rejected as compared with $10,000$; so we may round off the enlarged $y$ to $10,200$. $y$ has grown from $10,000$ to $10,200$; the bit added on is $dy$, which is therefore $200$.

${\frac {dy}{dx}}={\frac {200}{1}}=200$. According to the algebra-working of the previous paragraph, we find ${\frac {dy}{dx}}=2x$. And so it is; for $x=100$ and $2x=200$.

But, you will say, we neglected a whole unit.

Well, try again, making $dx$ a still smaller bit.

Try $dx={\tfrac {1}{10}}$. Then $x+dx=100.1$, and

$(x+dx)^{2}=100.1\times 100.1=10,020.01$.

Now the last figure $1$ is only one-millionth part of the $10,000$, and is utterly negligible; so we may take $10,020$ without the little decimal at the end. And this makes $dy=20$; and ${\frac {dy}{dx}}={\frac {20}{0.1}}=200$, which is still the same as $2x$.

Case 2.

Try differentiating $y=x^{3}$ in the same way.

We let $y$ grow to $y+dy$, while $x$ grows to $x+dx$.

Then we have

$y+dy=(x+dx)^{3}$.