Assuming that those who are reading have a minimum level in Maths, everyone knows perfectly that the quotient rule is #color(blue)(((u(x))/(v(x)))^'=(u^'(x)*v(x)-u(x)*v'(x))/((v(x))²))#, where #u(x)# and #v(x)# are functions and #u'(x)#, #v'(x)# respective derivates. But, where does it come from?
Let's find out!
Considering #f(x)=(u(x))/(v(x))#, and by definition, #f'(x)=lim_(h to 0)(f(x+h)-f(x))/h#
So :
#f'(x)=lim_(h to 0) ((u(x+h))/(v(x+h))-(u(x))/(v(x)))/h#
Now we need a common denominator :
#f'(x)=lim_(h to 0)((u(x+h)color(red)(*v(x)))/(v(x+h)color(red)(*v(x)))-(u(x)color(red)(*v(x+h)))/(v(x)color(red)(*v(x+h))))/h#
#f'(x)=lim_(h to 0)(u(x+h)color(red)(*v(x))-u(x)color(red)(*v(x+h)))/(v(x)*v(x+h))*1/h#
#f'(x)=lim_(h to 0)(u(x+h)color(red)(*v(x))-u(x)color(red)(*v(x+h)))/(h*v(x)*v(x+h))#
This expression isn't very useful for the moment, so let's add an intelligent 0:
#f'(x)=lim_(h to 0)(u(x+h)v(x)-u(x)v(x+h)color(red)(+u(x)v(x)-u(x)v(x)))/(hv(x)v(x+h))#
Now we can factoring :
#f'(x)=lim_(h to 0)(v(x)(u(x+h)-u(x))+u(x)(v(x)-v(x+h)))/(hv(x)v(x+h))#
Now we can cut our limit into two limits :
#f'(x)=lim_(h to 0)(v(x)(u(x+h)-u(x)))/(hv(x)v(x+h))+lim_(h to 0)(u(x)(v(x)-v(x+h)))/(hv(x)v(x+h))#
#f'(x)=lim_(h to 0)(v(x))/(v(x)v(x+h))*cancel((u(x+h)-u(x))/h)^(=u'(x))-lim_(h to 0)(u(x)(v(x+h)-v(x)))/(hv(x)v(x+h))#
#f'(x)=lim_(h to 0)(v(x)u'(x))/(v(x)(v(x+h)))-lim_(h to 0)(u(x))/(v(x)v(x+h))*cancel((v(x+h)-v(x))/h)^(=v'(x))#
#f'(x)=lim_(h to 0)(v(x)u'(x))/(v(x)(v(x+h)))-lim_(h to 0)(u(x)v'(x))/(v(x)v(x+h))#
#f'(x)=lim_(h to 0)(v(x)u'(x)-u(x)v'(x))/(v(x)v(x+h))#
And because #v(x+h)≈_(h to 0)v(x)#,
#f'(x)=(u^'(x)*v(x)-u(x)*v'(x))/((v(x))²)#
\0/ here's our answer !
