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: