Page:Calculus Made Easy.pdf/59

Lastly, we have to differentiate quotients.

Think of this example, ${\displaystyle y={\dfrac {bx^{5}+c}{x^{2}+a}}}$. In such a case it is no use to try to work out the division beforehand, because ${\displaystyle x^{2}+a}$ will not divide into ${\displaystyle bx^{5}+c}$, neither have they any common factor. So there is nothing for it but to go back to first principles, and find a rule.

So we will put ${\displaystyle y={\frac {u}{v}}}$;

where ${\displaystyle u}$ and ${\displaystyle v}$ are two different functions of the independent variable ${\displaystyle x}$. Then, when ${\displaystyle x}$ becomes ${\displaystyle x+dx}$, ${\displaystyle y}$ will become ${\displaystyle y+dy}$; and ${\displaystyle u}$ will become ${\displaystyle u+du}$; and ${\displaystyle v}$ will become ${\displaystyle v+dv}$. So then

Now perform the algebraic division, thus:

${\displaystyle v+dv\,\,}$	${\displaystyle u+du}$	${\displaystyle {\dfrac {u}{v}}+{\dfrac {du}{v}}-{\dfrac {u\cdot dv}{v^{2}}}}$
	${\displaystyle u+{\dfrac {u\cdot dv}{v}}}$
${\displaystyle du-{\dfrac {u\cdot dv}{v}}}$ ${\displaystyle du+{\dfrac {du\cdot dv}{v}}}$
${\displaystyle -{\dfrac {u\cdot dv}{v}}-{\dfrac {du\cdot dv}{v}}}$ ${\displaystyle -{\dfrac {u\cdot dv}{v}}-{\dfrac {u\cdot dv\cdot dv}{v^{2}}}}$
${\displaystyle -{\dfrac {du\cdot dv}{v}}+{\dfrac {u\cdot dv\cdot dv}{v^{2}}}}$.