Respuesta :
Presuming that the derivative that we are looking for is dy/dx.
Let's determine the derivative of the following equation:
[tex]\sin \mleft(x^{2}+y\mright)=x+y^2[/tex]We get,
[tex]\sin \mleft(x^{2}+y\mright)=x+y^2[/tex][tex]\sin \mleft(x^2+y\mright)-x-y^2=\text{ 0}[/tex][tex]\frac{d\text{y}}{dx}\lbrack\sin (x^2+y)-x-y^2\text{\rbrack}[/tex][tex]\frac{d\text{y}}{dx}\sin (x^2+y)-\frac{d\text{y}}{dx}x+\frac{d\text{y}}{dx}(-y^2)[/tex][tex]\cos (x^2+y)\cdot\frac{d\text{y}}{d\text{x}}(x^2\text{ + y) - 1 + 0}[/tex][tex]\cos (x^2+y)\cdot(\frac{d\text{y}}{d\text{x}}x^2\text{ + }\frac{d\text{y}}{d\text{x}}\text{y})\text{ - 1}[/tex][tex]\cos (x^2+y)\cdot(2x+0)\text{ - 1}[/tex][tex]2x\cos (x^2+y)-1[/tex]Therefore, the derivative (dy/dx) of sin(x² + y) = x + y^2 is:
[tex]2x\cos (x^2+y)-1[/tex]