Quotient Rule
Key Questions

#y'=(g(x)f'(x)f(x)g'(x))/(g(x))^2# #y=f(x)/g(x)=(2x^43x)/(4x1)# #f'(x)=8x^33# #g'(x)=4# #(g(x))^2=(4x1)^2# #y'=((4x1)(8x^33)(2x^43x)(4))/(4x1)^2# #y'=(32x^412x8x^3+38x^4+12x)/(4x1)^2# Simplify for combining like terms.
#Solution>y'=(24x^48x^3+3)/(4x1)^2# 
#y'=1/(sqrtx)*1/((1sqrtx)^2)# Explanation :
Using Quotient Rule, which is
#y=f(x)/g(x)# , then#y'=(g(x)f'(x)f(x)g'(x))/(g(x))^2# Similarly following for the given problem,
#y=(1+sqrtx)/(1sqrtx)# #y'=((1sqrtx)(1/(2sqrtx))(1+sqrtx)(1/(2sqrtx)))/((1sqrtx)^2)# #y'=1/(2sqrtx)*(1sqrtx+1+sqrtx)/((1sqrtx)^2)# #y'=1/(2sqrtx)*(2)/((1sqrtx)^2)# #y'=1/(sqrtx)*1/((1sqrtx)^2)# 
Answer:
#((u(x))/(v(x)))^'=(u^'(x)*v(x)u(x)*v'(x))/((v(x))²)# Explanation:
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 !