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

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