Question

x^3 = (2x-y)/(x+3y)
solve for dy/dx

Answer 1

\[x ^{3}\left( x+3y \right)=2x-y,x ^{4}+3x ^{3}y=2x-y,3x^3y+y=2x-x ^{4}\]
\[y \left( 3x^3+1 \right)=2x-x^4\]
\[y=\frac{ 2x-x^4 }{ 3x^3+1 }\]
\[\frac{ dy }{dx }=\frac{ \left( 3x^3+1 \right)\left( 2-4x^3 \right)-\left( 2x-x^4 \right)\left( 6x \right) }{ \left( 3x^3+1 \right)^2}\]
simplify it.

Answer 2

write 9x^2 in place of 6x

Answer 3

wow that is way off than what I thought

Answer 4

okay so far I have -16x^3 - 15 x^6 +2 / (3x^2 +1)^2