Question

a expressão derivada y' para a função y=f(x) dada implicitamente pela expressão x^3-3x^2y^4+4y^3=6x+1 é:

Answer 1

Bom Dia. Seguiremos as seguintes etapas:

$x^{3}-3x^{2}y^{4}+4y^{3}=6x+1 \\ \\ 3x^{2}-(3x^{2}'y^{4}+3x^{2}y^{4}')+12y^{2} \displaystyle \frac{dy}{dx} =6 \\ \\ 3x^{2}-(6xy^{4}+3x^{2}4y^{3}\displaystyle \frac{dy}{dx})+12y^{2} \displaystyle \frac{dy}{dx} =6 \\ \\ 3x^{2}-(6xy^{4}+12x^{2}y^{3}\displaystyle \frac{dy}{dx})+12y^{2} \displaystyle \frac{dy}{dx} =6 \\ \\ 3x^{2}-6xy^{4}-12x^{2}y^{3}\displaystyle \frac{dy}{dx}+12y^{2} \displaystyle \frac{dy}{dx} =6$

$-12x^{2}y^{3}\displaystyle \frac{dy}{dx}+12y^{2} \displaystyle \frac{dy}{dx} =-3x^{2}+6xy^{4}+6 \\ \\ \displaystyle \frac{dy}{dx}(-12x^{2}y^{3}+12y^{2})=-3x^{2}+6xy^{4}+6 \\ \\ \displaystyle \frac{dy}{dx}= \displaystyle \frac{-3x^{2}+6xy^{4}+6}{-12x^{2}y^{3}+12y^{2}} \\ \\ \displaystyle \frac{dy}{dx}= \displaystyle \frac{6xy^{4}-3x^{2}+6}{12y^{2}-12x^{2}y^{3}} \\ \\ \displaystyle \frac{dy}{dx}= \displaystyle \frac{3(2xy^{4}-x^{2}+2)}{3(4y^{2}-4x^{2}y^{3})}$

$\boxed{\displaystyle \frac{dy}{dx}= \displaystyle \frac{2xy^{4}-x^{2}+2}{4y^{2}-4x^{2}y^{3}}}$