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