Question

Derivacao implícita
2x^3+x^2vezes y+y^3=1

Answer 1

$\frac{\mathrm{d} }{\mathrm{d} x}2x^3+x^2 y+y^3=\frac{\mathrm{d} }{\mathrm{d} x}1 \\ =6x^2+2xy+x^2\frac{\mathrm{d}y }{\mathrm{d} x}+3y^2\frac{\mathrm{d} y }{\mathrm{d} x}=0 =\ \textgreater \ \\ (x^2+3y^2)\frac{\mathrm{d} y}{\mathrm{d} x} = -(6x^2+2xy)\\ \frac{\mathrm{d}y }{\mathrm{d} x}= \frac{-(6x^2+2xy)}{(x^2+3y^2)}$