Page:Calculus Made Easy.pdf/59

Lastly, we have to differentiate quotients.

Think of this example, $y={\dfrac {bx^{5}+c}{x^{2}+a}}$. In such a case it is no use to try to work out the division beforehand, because $x^{2}+a$ will not divide into $bx^{5}+c$, 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 $y={\frac {u}{v}}$;

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

$y+dy={\dfrac {u+du}{v+dv}}$.

Now perform the algebraic division, thus:

$v+dv\,\,$

$u+du$

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

$u+{\dfrac {u\cdot dv}{v}}$

$du-{\dfrac {u\cdot dv}{v}}$

$du+{\dfrac {du\cdot dv}{v}}$

	$-{\dfrac {u\cdot dv}{v}}-{\dfrac {du\cdot dv}{v}}$
	$-{\dfrac {u\cdot dv}{v}}-{\dfrac {u\cdot dv\cdot dv}{v^{2}}}$

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