Respuesta :
Answer:
dx/dy = (x^4 - 2x^5y - 6xy^2) / (5x^4y^2 - 4x^3y + 2y^3).
Step-by-step explanation:
x^5y^2 − x^4y + 2xy^3 = 0
Applying the Product and Chain Rules:
y^2*5x^4*dx/dy + 2y*x^5 - (y*4x^3*dx/dy + x^4) + (y^3* 2*dx/dy + 3y^2*2x) =0
Separating the terms with derivatives:
y^2*5x^4*dx/dy - y*4x^3*dx/dy + y^3* 2*dx/dy = x^4 - 2y*x^5 - 3y^2*2x
dx/dy = (x^4 - 2x^5y - 6xy^2) / (5x^4y^2 - 4x^3y + 2y^3)
Answer:
[tex]\frac{dx}{dy}=\frac{-2x^5y+x^4-6xy^2}{5x^4y^2-4x^3y+2y^3}[/tex]
Step-by-step explanation:
The given equation is
[tex]x^5y^2-x^4y+2xy^3=0[/tex]
Differentiate with respect to y.
[tex]\frac{d}{dy}(x^5y^2)-\frac{d}{dy}(x^4y)+\frac{d}{dy}(2xy^3)=0[/tex]
Using product rule we get
[tex]x^5\frac{d}{dy}(y^2)+y^2\frac{d}{dy}(x^5)-x^4\frac{d}{dy}(y)-y\frac{d}{dy}(x^4)+2x\frac{d}{dy}(y^3)+2y^3\frac{d}{dy}(x)=0[/tex] [tex](fg)'=fg'+gf'[/tex]
[tex]x^5(2y)+y^2(5x^4\frac{dx}{dy})-x^4(1)-y(4x^3\frac{dx}{dy})+2x(3y^2)+2y^3\frac{dx}{dy}=0[/tex]
[tex]2x^5y+5x^4y^2\frac{dx}{dy}-x^4-4x^3y\frac{dx}{dy}+6xy^2+2y^3\frac{dx}{dy}=0[/tex]
Isolate [tex]\frac{dx}{dy}[/tex] terms on left side.
[tex]5x^4y^2\frac{dx}{dy}-4x^3y\frac{dx}{dy}+2y^3\frac{dx}{dy}=-2x^5y+x^4-6xy^2[/tex]
[tex](5x^4y^2-4x^3y+2y^3)\frac{dx}{dy}=-2x^5y+x^4-6xy^2[/tex]
Isolate [tex]\frac{dx}{dy}[/tex] term.
[tex]\frac{dx}{dy}=\frac{-2x^5y+x^4-6xy^2}{5x^4y^2-4x^3y+2y^3}[/tex]
Therefore the value of [tex]\frac{dx}{dy}[/tex] is [tex]\frac{-2x^5y+x^4-6xy^2}{5x^4y^2-4x^3y+2y^3}[/tex].
