How do you differentiate $3xy^2+cosy^2=2x^3+5$ ?

Question

4192 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-12T05:46:03

$dy/dx=(3(2x^2-y^2))/(2y(3x-siny^2))$

$d/dx(3xy^2+cosy^2)=d/dx(2x^3+5)$

$d/dx3xy^2+d/dxcosy^2=d/dx2x^3+d/dx5$

$[d/dx3xy^2]+d/dxcosy^2=d/dx2x^3+d/dx5$

$[3x2ydy/dx+y^2 3]+(-2ysiny^2)dy/dx=6x^2+0$
differentiate of $siny^2$ and $sin^2y$ is different answer

$6xydy/dx+3y^2 -(2ysiny^2) dy/dx=6x^2$

$dy/dx(6xy -2ysiny^2) =6x^2-3y^2$
$dy/dx=(6x^2-3y^2)/(6xy -2ysiny^2)=(3(2x^2-y^2))/(2y(3x-siny^2))$

How do you differentiate 3xy^2+cosy^2=2x^3+53xy^2+cosy^2=2x^3+5?