Find dy/dx for the equation x^3-2x^2y+4xy^2=20
its instructed to use the the product rule for the middle terms which is where things become fuzzy I have 3x^2(product rule)=0 I really could use some help

Question

Find dy/dx for the equation x^3-2x^2y+4xy^2=20
its instructed to use the the product rule for the middle terms which is where things become fuzzy I have 3x^2(product rule)=0 I really could use some help

anonymous · Answer

Considering $$\frac{dy}{dx}=y'$$$$x^3-2x^2y+4xy^2=20$$Let's make the derivative for each member
Exponential rule$$x^3=3x^2$$Product rule$$-2x^2y=-2*(x^2y'+2yx)$$Product rule again$$4xy^2=4*(2xyy'+y^2)$$Derivativo of constant$$20=0$$Pull everything together$$3x^2-2x^2y'-4yx+8xyy'+4y^2=0$$$$y'*(8xy-2x^2)=4xy-3x^2+4y^2$$The answer is:$$y'=\frac{4xy-3x^2+4y^2}{(8xy-2x^2)}$$

anonymous · Answer

thank you!