Page:Calculus Made Easy.pdf/60

 As both these remainders are small quantities of the second order, they may be neglected, and the division may stop here, since any further remainders would be of still smaller magnitudes.

So we have got: $y+dy=\frac{u}{v}+\frac {du}{v}-\frac{u\cdot dv}{v^2}$; which may be written $=\frac{u}{v}+\frac{v\cdot du-u\cdot dv}{v^2}$. Now subtract the original $$y=\frac{u}{v}$$, and we have left: $dy=\frac{v\cdot du-u\cdot dv}{v^2}$; whence $\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$.|undefined

This gives us our instructions as to ''how to differentiate a quotient of two functions. Multiply the divisor function by the differential coefficient of the dividend function; then multiply the dividend function by the differential coefficient of the divisor function; and subtract. Lastly divide by the square of the divisor function.''

Going back to our example $$y=\frac{bx^5+c}{x^2+a}$$,

write $bx^5+c=u$; and $x^2+a=v$.