How do you implicitly differentiate #-1=y^3x-2xy-9x^4y #?

1 Answer

#y'=(y^3-2y-36x^3y)/(9x^4+2x-3x*y^2)#

Explanation:

From the given equation we differentiate both sides with respect to x

#-1=y^3*x-2xy-9x^4*y#

#d/dx(-1)=d/dx(y^3*x-2xy-9x^4*y)#

#d/dx(-1)=d/dx(y^3*x)-2*d/dx(xy)-9*d/dx(x^4*y)#

#0=y^3*d/dx(x)+x*d/dx(y^3)-2[x*d/dx(y)+y*d/dx(x)]-9[x^4d/dx(y)+yd/dx(x^4)]#

#0=y^3+x*(3y^2*y')-2[x*(y')+y*1]-9[x^4(y')+y(4x^3)]#

#0=y^3+3x*y^2*y'-2xy'-2y-9x^4y'-36x^3y#

transposition to obtain

#9x^4y'+2xy'-3x*y^2*y'=y^3-2y-36x^3y#

factoring to obtain

#(9x^4+2x-3x*y^2)y'=y^3-2y-36x^3y#

divide both sides of the equation by #(9x^4+2x-3x*y^2)#

#((9x^4+2x-3x*y^2)y')/(9x^4+2x-3x*y^2)=(y^3-2y-36x^3y)/(9x^4+2x-3x*y^2)#

#(cancel((9x^4+2x-3x*y^2))y')/cancel((9x^4+2x-3x*y^2))=(y^3-2y-36x^3y)/(9x^4+2x-3x*y^2)#

#y'=(y^3-2y-36x^3y)/(9x^4+2x-3x*y^2)#

God bless...I hope the explanation is useful.