Find (dy)/(dx) when \tt {x}^(2) + {y}^(2) = sin \: xy Please help! :) Thanks in advance!! ​

Question

Find
`(dy)/(dx)` when
`\tt {x}^(2) + {y}^(2) = sin \: xy`

Please help! :)
Thanks in advance!! ​

Answer 1

`\boxed{\bold{ \tt (dy)/(dx)=(2x-y cos\: xy)/(x*cos \:xy-2y)}}`

Explanation:

`\tt x^2+y^2=sin xy`

Differentiating both sides with respect to x.

`\tt{(d)/(dx)(x^2+y^2)=(d)/(dx)(sin xy)}`

Using the Addition rule, Power rule, chain rule, and Product rule respectively.

`\bold{\tt (d)/(dx)x^2+(d)/(dx){y^2}= (d sin xy)/(dxy)*(dxy)/(dx)}`

`\bold{\tt2x^(2-1)+(d y^2)/(dx)*(dy)/(dx)=cosxy*(y*(dx)/(dy)+x(dy)/(dy)*(dy)/(dx))}`

`\bold{ \tt2x+2y(dy)/(dx)= cos xy*(y+x*(dy)/(dx))}`

`\bold{ \tt2x+2y(dy)/(dx)=y cos\: xy+x(dy)/(dx)*cos \:xy}`

Solving for
`\tt (dy)/(dx)`

`\bold{ \tt2x-y cos\: xy=x(dy)/(dx)*cos \:xy-2y(dy)/(dx)}`

`\bold{ \tt x(dy)/(dx)*cos \:xy-2y(dy)/(dx)=2x-y cos\: xy}`

Taking common
`\tt (dy)/(dx)`

`\bold{ \tt (dy)/(dx)(x*cos \:xy-2y)=2x-y cos\: xy}`

Solving for
`\tt (dy)/(dx)`

`\bold{ \tt (dy)/(dx)=(2x-y cos\: xy)/(x*cos \:xy-2y)}`

Therefore, Answer is
`\boxed{\bold{ \tt (dy)/(dx)=(2x-y cos\: xy)/(x*cos \:xy-2y)}}`

Note: Formula

`\boxed{\bold{\tt{Addition \: Rule:(d)/(dx)(x^n+y^n) =(d)/(dx)*x^n+(d)/(dx)*y^n}}}`

`\boxed{\bold{\tt{Power \: Rule:(d)/(dx)x^n =n*x^(n-1)}}}`

`\boxed{\bold{\tt{Chain \:\: Rule: (d)/(dx)y^n=(d)/(dy)y^n(dy)/(dx)=n*y^(n-1)(dy)/(dx)}}}`

`\boxed{\bold{\tt{Product\:Rule:(d)/(dx)(u*v)=(du)/(dx)*v+u*(dv)/(dx)}}}`