Question

Find the derivative sin(x² + y) = x + y^2

Answer 1

Presuming that the derivative that we are looking for is dy/dx.

Let's determine the derivative of the following equation:

`\sin \mleft(x^(2)+y\mright)=x+y^2`

We get,

`\sin \mleft(x^(2)+y\mright)=x+y^2`
`\sin \mleft(x^2+y\mright)-x-y^2=\text{ 0}`
`\frac{d\text{y}}{dx}\lbrack\sin (x^2+y)-x-y^2\text{\rbrack}`
`\frac{d\text{y}}{dx}\sin (x^2+y)-\frac{d\text{y}}{dx}x+\frac{d\text{y}}{dx}(-y^2)`
`\cos (x^2+y)\cdot\frac{d\text{y}}{d\text{x}}(x^2\text{ + y) - 1 + 0}`
`\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}`
`\cos (x^2+y)\cdot(2x+0)\text{ - 1}`
`2x\cos (x^2+y)-1`

Therefore, the derivative (dy/dx) of sin(x² + y) = x + y^2 is:

`2x\cos (x^2+y)-1`