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