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STEP 1: $$ (x+y)^{3} = x^3 + y^3 $$ STEP 2: $$ 3(x+y)^2 (1+ dy/dx) = 3x^2 + 3y^2(dy/dx) $$ STEP 3: $$ 3(x+y)^2 + \frac{dy}{dx}\cdot3(x+y)^2 = 3x^2 + 3y^2(dy/dx)$$ STEP 4: $$\frac{dy}{dx} \cdot 3(x+y)^2 = 3x^2 + 3y^2(dy/dx) - 3(x+y)^2$$ STEP 5: $$\frac{dy}{dx} = (3x^2 + 3y^2(dy/dx))/3(x+y)^2 - 1$$ I do not quite get how to move the dy/dx to one side, and all the others to the other side to solve for what dy/dx is.

It is one of my worst weaknesses when I try to do an implicit differentiation. I've searched thorugh wolfram alpha to see how such simplification works, but some of the problems seems to only have simple step-by-step process shown, not the whole step-by-step solution.

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  • $\begingroup$ You missed a $\frac{dy}{dx}$ on the right side in step $4$. $\endgroup$ – Mike Sep 22 '14 at 21:55
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From here it should be algebra. First, clear the parentheses on that $1+\frac{dy}{dx}$ term. Once that's done, you should be able to collect all terms with $\frac{dy}{dx}$ on one side of the equation and the rest on the other. Then simply divide to get $\frac{dy}{dx}$ alone.

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Hint $x^3 + y^3 = (x+y)(x^2-xy+y^2)$

so we have the expression

$$(x+y)^3=(x+y)(x^2-xy+y^2)$$

$$(x+y)[(x+y)^2-(x^2-xy+y^2)]=0$$

$$(x+y)(3xy)=0$$

$$x^2y+y^2x=0$$

which when differentiated becomes

$$x^2\displaystyle\frac{dy}{dx}+y(1)+y^2(1)+x\displaystyle\frac{dy}{dx}=0$$

$$\displaystyle\frac{dy}{dx}(x^2+x)=-(y^2+y)$$

$$\displaystyle\frac{dy}{dx}=-\displaystyle\frac{y^2+y}{x^2+x}$$

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