Page:Calculus Made Easy.pdf/40

Numerical example.

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

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

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

Well, try again, making ${\displaystyle dx}$ a still smaller bit.

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

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

Case 2.

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

We let ${\displaystyle y}$ grow to ${\displaystyle y+dy}$, while ${\displaystyle x}$ grows to ${\displaystyle x+dx}$.

Then we have