Question

Find dy/dx by implicit differentiation. for: x^4(x+y) = y^2(3x-y)

Answer 1

\[\frac{d}{dx}(x^4(x+y))=\frac{d}{dx}(y^2(3x-y))\]For the left-hand side:\[\rightarrow x^4\frac{d}{dx}(x+y)+(x+y)\frac{d}{dx}x^4=x^4(1+\frac{dy}{dx})+(x+y)4x^3\]\[=x^4\frac{dy}{dx}+5x^4+4x^3y\]and for the right-hand side,\[y^2\frac{d}{dx}(3x-y)+(3x-y)\frac{d}{dx}y^2=y^2(3-\frac{dy}{dx})+(3x-y)2y \frac{dy}{dx}\]\[=3y^2+3y(2x-y)\frac{dy}{dx}\]Hence, equating LHS and RHS:\[x^4\frac{dy}{dx}+5x^4+4x^3y=3y^2+3y(2x-y)\frac{dy}{dx}\]

Answer 2

Now you can solve for dy/dx:\[\frac{dy}{dx}=\frac{3y^2-5x^4-4x^3y}{x^4-6xy-3y^2}\]