Question

Suppose y = (2 x^3 - 1)/(4 x^4+ 1). Then

(dy)/(dx) = (-8 x^A + B x^3 + C x^D)/(4 x^4 + 1)^F

what is abcdf?

Answer 1

i found a = 6 f = 2

Answer 2

So if :
\[y=\frac{2x^3-1}{4x^4+1}\]
and you want to differentiate it then you get:
\[\frac{dy}{dx}=\frac{(6x^2)(4x^4+1)-(2x^3-1)(16x^3)}{(4x^4+1)^2}=\frac{2x^2(12x^4+3-16x^4+8x)}{(4x^4+1)^2}=\]
\[\frac{-2x^2(4x^4-8x-3)}{(4x^4+1)^2}=\frac{-8x^6+16x^3+6x^2}{(4x^4+1) ^2} \]
A=6,B=16,C=6, D=2,F=2
So