Question

derivar a função x²/(x+y ) = y² + 1 *

Answer 1

Temos a seguinte equação:

$\sf \frac{x {}^{2} }{(x + y)} = y {}^{2} + 1$

Aplicando a derivação implícita:

$\sf \frac{d}{dx} \left( \frac{x {}^{2} }{x + y} \right) = \frac{d}{dx} (y {}^{2} + 4)$

Aplicando a regra do quociente no primeiro membro e no segundo a derivada da soma;

$\sf \frac{ \frac{d}{dx}(x {}^{2} ).(x + y) - x {}^{2} . \frac{d}{dx} (x + y)}{(x + y) {}^{2} } = \frac{d}{dx} y {}^{2} + \frac{d}{dx}4 \\ \\ \sf \frac{2x.(x + y) - x {}^{2}.(1 + 1. \frac{dy}{dx} )}{(x + y) {}^{2} } = 2y. \frac{dy}{dx} + 0 \\ \\ \sf \frac{2x {}^{2} +2xy - x {}^{2} - x {}^{2}. \frac{dy}{dx} }{(x + y) {}^{2} } = 2y. \frac{dy}{dx} \\ \\ \sf \sf 2{x}^{2} + 2xy - x {}^{2} - x {}^{2} . \frac{dy}{dx} = 2y. \frac{dy}{dx} .(x + y){}^{2} \\ \\ \sf 2 {x}^{2} + 2xy - x {}^{2} - x {}^{2} . \frac{dy}{dx} = 2y. \frac{dy}{dx} .(x {}^{2} + 2xy + y {}^{2} ) \\ \\ \sf 2x {}^{2} + 2xy - x {}^{2} - x {}^{2} . \frac{dy}{dx} = 2yx {}^{2} . \frac{dy}{dx} + 4y {}^{2} x. \frac{dy}{dx} + 2y {}^{3} . \frac{dy}{dx} \\ \\ \sf 2x {}^{2} + 2xy - x {}^{2} = 2yx {}^{2} . \frac{dy}{dx} + 4y {}^{2} x. \frac{dy}{dx} + 2y {}^{3} . \frac{dy}{dx} + x {}^{2} . \frac{dy}{dx} \\ \\ \sf 2x {}^{2} + 2xy - x {}^{2} = \frac{dy}{dx} .(2y {x}^{2} + 4y {}^{2} x + 2y {}^{3} + x {}^{2} ) \\ \\ \boxed{ \boxed{ \boxed{ \boxed{ \sf \frac{dy}{dx} = \frac{x {}^{2} + 2xy }{2yx {}^{2} + 4y {}^{2}x + 2y {}^{3} + x {}^{2} } }}}}$

Espero ter ajudado