How do you find the derivative of #y=(x^3+x)/(4x+1)#?

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Answer 1 · 2015-12-25T22:21:47

#y=(8x^3+3x^2+1)/(4x+1)^2#

If #y=(f(x))/(g(x))#, then

#y'=(f'(x)g(x)-g'(x)f(x))/(g(x))^2#

So,

#f(x)=x^3+x#
#g(x)=4x+1#

#f'(x)=3x^2+1#
#g'(x)=4#

Plug these in:

#y'=((3x^2+1)(4x+1)-4(x^3+x))/(4x+1)^2#

#=>(12x^3+3x^2+4x+1-4x^3-4x)/(4x+1)^2#

#=>(8x^3+3x^2+1)/(4x+1)^2#