Quotient Rule
Key Questions
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#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# -
#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)# -
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 !