Page:Calculus Made Easy.pdf/57

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:

${\frac {dy}{dx}}=6ax^{5}+4acx^{3}+2bx$.

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:

$y+dy=(u+du)\times (v+dv)$

$=u\cdot v+u\cdot dv+v\cdot du+du\cdot dv$.

Now $du\cdot dv$ is a small quantity of the second order of smallness, and therefore in the limit may be discarded, leaving

$y+dy=u\cdot v+u\cdot dv+v\cdot du$.