Using implicit differentiation of multiple varibles

Question

The right-hand side of the equation was easier to differentiate wrt $x$ in which I got:

$4x-4y(\frac{dy}{dx})$

However, that was not the case for the left-hand side.

The main reason is the term is inside the square, so I cannot directly take its derivative. Expanding it was one option, however, this is only useful for low exponents. Is there a much more simpler way to differentiate the left-hand side?

Answer 1

From $(x^2 + y^2)^2 = 2(x^2 -y^2)$,

we have: $2(x^2 + y^2)(2x + 2yy') = 4(x - yy')$

Then: $(x^2 + y^2)(x + yy') = x - yy'$

$\Rightarrow$ $x(x^2 + y^2) + yy'(x^2 + y^2) = x - yy'$

$\Rightarrow$ $x(x^2 + y^2 - 1) = - yy'(x^2 + y^2 + 1)$

$\Rightarrow$ $y' = \dfrac{x(x^2 + y^2 - 1)}{- y(x^2 + y^2 + 1)}$

Therefore, $\dfrac{dy}{dx} = y' = \dfrac{x(x^2 + y^2 - 1)}{- y(x^2 + y^2 + 1)}$

Answer 2

The chain rule still applies to implicit differentiation. Set $u=x^2+y^2$, and the derivative of the left-hand side becomes $$ \frac{d}{dx}u^2=2u\cdot\frac{du}{dx}\\ =2(x^2+y^2)\cdot\frac{d}{dx}(x^2+y^2) $$ You managed to differentiate the right-hand side of the original equation, so differentiating $x^2+y^2$ to complete this shouldn't be too different.