Question

If `u = arcsin( (x^2+y^2) / (x+y) )` then does `x(partial u)/(partial x) + y(partial u)/(partial y) = tan u`?

Answer 1

The expression is valid.

Explanation:

We have:

`u = arcsin( (x^2+y^2) / (x+y) )`

Using the known result:

`d/dx arcsinx = 1/sqrt(1-x^2)`

In conjunction with the chain rule and quotient rule, we can calculate the Partial Derivative wrt `x`:

`(partial u)/(partial x) = 1/sqrt( 1- ( (x^2+y^2) / (x+y) )^2) \ (partial)/(partial x) (x^2+y^2) / (x+y)`

`\ \ \ \ \ \ = 1/sqrt( 1- ( (x^2+y^2) / (x+y) )^2) \ (partial)/(partial x) (x^2+y^2) / (x+y)`

`\ \ \ \ \ \ = 1/sqrt( 1- ( (x^2+y^2) / (x+y) )^2) \ ( (x+y)(2x) - (x^2+y^2)(1) ) / (x+y)^2`

`\ \ \ \ \ \ = 1/sqrt( 1- ( (x^2+y^2) / (x+y) )^2) \ ( 2x^2+2xy - x^2 - y^2 ) / (x+y)^2`

`\ \ \ \ \ \ = ( x^2+2xy - y^2 )/((x+y)^2sqrt( 1- ( (x^2+y^2) / (x+y) )^2))`

Similarly:

`(partial u)/(partial y) = ( y^2+2xy - x^2 )/((x+y)^2sqrt( 1- ( (x^2+y^2) / (x+y) )^2))`

And so we compute the LHS of the required result:

`x(partial u)/(partial x) + y(partial u)/(partial y)`

`= ( x(x^2+2xy - y^2) )/((x+y)^2sqrt( 1- ( (x^2+y^2) / (x+y) )^2)) + ( y(y^2+2xy - x^2) )/((x+y)^2sqrt( 1- ( (x^2+y^2) / (x+y) )^2))`

`= ( x^3+2x^2y -x y^2 + y^3+2xy^2 - x^2y )/((x+y)^2sqrt( 1- ( (x^2+y^2) / (x+y) )^2))`

`= ( x^3+x^2y + y^3+xy^2 )/((x+y)^2sqrt( 1- ( (x^2+y^2) / (x+y) )^2))`

`= ( (x+y)(x^2+y^2) )/((x+y)^2sqrt( 1- ( (x^2+y^2) / (x+y) )^2))`

`= ( (x^2+y^2) )/((x+y)sqrt( 1- ( (x^2+y^2) / (x+y) )^2))` ...[A]

We now seek an expression for `tanu`:

`cos^2u = 1 - ( (x^2+y^2)/(x+y))^2 => cosu = sqrt(1 - ( (x^2+y^2)/(x+y))^2)`

And by definition:

`tan u = sinu/cosu`
`\ \ \ \ \ \ \ \ \= { (x^2+y^2)/(x+y) } / {sqrt(1 - ( (x^2+y^2)/(x+y))^2)}`

`\ \ \ \ \ \ \ \ \= { (x^2+y^2) } / {(x+y)sqrt(1 - ( (x^2+y^2)/(x+y))^2)}`

Which is the same as [A], and thus we confirm that

`x(partial u)/(partial x) + y(partial u)/(partial y) = tan u`