How do you differentiate #f(x)=(xtanx)/(x^2-x+3)# using the quotient rule?

1 Answer
Mar 29, 2018

#-x*(x^2-x+3)^-2*(2x-1)*tanx +(x^2-x+3)^-1*tanx+x*(x^2-x+3)^-1*sec^2x#

There are 2 Answers first with the product rule the second with the quotient rule

Explanation:

First Answer:
you can write the function like this: #f(x)=x*(x^2-x+3)^-1*tanx#

now consider:
#g(x)=x #

#h(x)=tanx#
#v(x)=(x^2-x+3)^-1#

#d/dx(g(x)*h(x)*v(x))=g'(h)*h(x)*v(x) +g(x)*h(x)*v'(x)+g(x)*h'(x)*v(x)#

#g'(x)=1#

#h'(x)=sec^2x#

#v'(x)=-(x^2-x+3)^-2*(2x-1)#

and by substituting in the relation above you get:
#d/dxx*(x^2-x+3)^-1*tanx=-x*(x^2-x+3)^-2*(2x-1)*tanx +(x^2-x+3)^-1*tanx+x*(x^2-x+3)^-1*sec^2x#

Second answer
#y'=(g(x)f'(x)−f(x)g'(x))/(g(x))^2#
where #f(x)=xtanx#
#g(x)=x^2-x+3#

and by differentiating them you get :
#f'(x)=xsec^2x+tanx#
#g'(x)=2x-1#
and by substituting in the above equation you get:
#y'=((x^2-x+3)(xsec^2x+tanx)-(xtanx)(2x-1))/(x^2-x+3)^2#