How do you find the derivative of #f(x)=(2x+6)/(3x^2+9)#?

2 Answers
Mar 21, 2016

#f'(x)=(-6(x^2+6x-3))/(3x^2+9)^2#

Explanation:

We should use quotient rule i.e if #f(x)=g(x)/(h(x))#, then

#f'(x)=(h(x)xxg'(x)-g(x)xxh'(x))/(h(x))^2#

As #g(x)=2x+6# and #h(x)=3x^2+9#

Hence #f'(x)=((3x^2+9)xx2-(2x+6)xx6x)/(3x^2+9)^2# or

#f'(x)=(6x^2+18-12x^2-36x)/(3x^2+9)^2# or

#f'(x)=((-6x^2-36x+18))/(3x^2+9)^2# or

#f'(x)=(-6(x^2+6x-3))/(3x^2+9)^2# or

#f'(x)=(-2(x^2+6x-3))/(3(x^2+3)^2)#

Mar 21, 2016

I get#" " (dy)/(dx)=- (2(x^2+6x-3))/(3(x^2+3)^2)#

Explanation:

Quotient rule states that for #y=u/v# where #u=f(x)" and "v=g(x)#

Then #dy/dx= (v (du)/(dx) - u (dv)/(dx))/v^2#

Set #y=f(x)=(2x+6)/(3x^2+9) =u/v#

Then #(du)/(dx)=2" "(dv)/(dx)=6x#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(brown)(=>dy/dxcolor(green)(= (v (du)/(dx) - u (dv)/(dx))/v^2) -> ((3x^2+9)(2)-(2x+6)(6x))/((3x^2+9)^2))#

Not that :#(3x^2+9)^2 = (3(x^2+3))^2 = 9(x^2+3)^2#

#" "=(2cancel((3x^2+9)))/((3x^2+9)^(cancel(2))) -(2(x+3)(6x))/(9(x^2+3)^2) #

#" "=2/(3(x^3+3))- (4x(x+3))/(3(x^3+3)^2)#

#" "=(2(x^2+3)-4x(x+3))/(3(x^2+3)^2)#

#" "=(2x^2+6-4x^2-12x)/(3(x^2+3)^2)#

#" "=(-2x^2-12x+6)/(3(x^2+3)^2)#

#" "(dy)/(dx)= (-2(x^2+6x-3))/(3(x^2+3)^2)#

#" "color(blue)((dy)/(dx)=- (2(x^2+6x-3))/(3(x^2+3)^2))#