Quotient Rule

Key Questions

  • How do you use the quotient rule to differentiate #y=(2x^4-3x)/(4x-1)#?

    Quotient Rule:

    #y'=(g(x)f'(x)-f(x)g'(x))/(g(x))^2#

    #y=f(x)/g(x)=(2x^4-3x)/(4x-1)#

    #f'(x)=8x^3-3#

    #g'(x)=4#

    #(g(x))^2=(4x-1)^2#

    #y'=((4x-1)(8x^3-3)-(2x^4-3x)(4))/(4x-1)^2#

    #y'=(32x^4-12x-8x^3+3-8x^4+12x)/(4x-1)^2#

    Simplify for combining like terms.

    #Solution->y'=(24x^4-8x^3+3)/(4x-1)^2#

    AJ Speller · 3 · Sep 23 2014
  • How do you use the quotient rule to find the derivative of #y=(1+sqrt(x))/(1-sqrt(x))# ?

    #y'=1/(sqrtx)*1/((1-sqrtx)^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)/(1-sqrtx)#

    #y'=((1-sqrtx)(1/(2sqrtx))-(1+sqrtx)(-1/(2sqrtx)))/((1-sqrtx)^2)#

    #y'=1/(2sqrtx)*(1-sqrtx+1+sqrtx)/((1-sqrtx)^2)#

    #y'=1/(2sqrtx)*(2)/((1-sqrtx)^2)#

    #y'=1/(sqrtx)*1/((1-sqrtx)^2)#

    Gaurav · · Sep 2 2014
  • What is the quotient rule?

    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 !

    Guillaume L. · 2 · May 30 2018

Questions

Basic Differentiation Rules