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