Given a implicit function tried to find the $\frac{du}{dx}$ and $\frac{dv}{dx}$ of $u(x,y)$ and $v(x,y)$ $$ x + y + u + v = a, x^3 + y^3+u^3+v^3 =b $$ I differentiated these equations: $$dx +dy + du + dv = 0, 3x^2dx+3y^2dy+3u^2du ++3v^2dv = 0$$

$$\Rightarrow $$ $$du = - \frac{1}{u^2} \left (v^2+y^2+x^2 \right ) ;dv = - \frac{1}{v^2} \left (u^2+y^2+x^2 \right )$$ $$\Rightarrow $$ $$\frac{dv}{dx} = -\frac{x^2}{v^2};\frac{du}{dx} = -\frac{x^2}{u^2}$$

but actually its wrong and correct answer is $\frac{du}{dx} = -\frac{v^2-x^2}{v^2-u^2}$ and $\frac{dv}{dx} = \frac{u^2-x^2}{v^2- u^2}$ So, I want to know where am I wrong and how to find differentials correctly?


Since $u$ and $v$ are functions of $x$ and $y$, then instead of simply $du$, you would have $\frac{du}{dx}dx+\frac{du}{dy}dy$, and similarly for $dv$.

While this approach can get the answer, another approach would give it to you easier. First, do derivative with respect to $x$. At this point, you could arrange your results into a system of equations, with $\frac{du}{dx}$ and $\frac{dv}{dx}$ as variables. You would solve this system to get the values of those derivatives. Then, you can repeat the process for derivative with respect to $y$.

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