Question

determine dy/dx implicitamente se y² = x² + ysenx .

Answer 1

Olá!

$y^2=x^2+y\cdot\sin\,x\\\\2ydy=2xdx+dy\cdot\sin\,x+y\cdot\cos\,xdx\\\\2ydy-\sin\,xdy=2xdx+y\cdot\cos\,xdx\\\\(2y-\sin\,x)dy=(2x+y\cdot\cos\,x)dx\\\\\boxed{\frac{dy}{dx}=\frac{2x+y\cdot\cos\,x}{2y-\sin\,x}}$

Answer 2

$\mathsf{y^2=x^2+ysen(x)}\\\mathsf{2y\dfrac{dy}{dx}=2x+\dfrac{dy}{dx}\cdot sen(x)+ycos(x)}\\\mathsf{2y\dfrac{dy}{dx}-\dfrac{dy}{dx}\cdot sen(x)=-2x-ycos(x)}\\\mathsf{\dfrac{dy}{dx}(2y-sen(x))=-(2x+ycos(x))}\\\large\boxed{\boxed{\boxed{\boxed{\mathsf{\dfrac{dy}{dx}=-\dfrac{2x+ycos(x)}{2y-sen(x)}}}}}}$