Question

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

Answer 1

`-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`