97.1k views
2 votes
If u=sin^(-1)(x^(2)+y^(2))/(x+y) then x(delu )/(delx )+y(delu )/(dely ) is

User Hellen
by
8.5k points

1 Answer

3 votes

Final answer:

The question asks to find the expression x(\partial u/\partial x) + y(\partial u/\partial y) for a given function u involving inverse sine. The solution requires applying the chain rule to differentiate u with respect to x and y, and then summing the weighted partial derivatives.

Step-by-step explanation:

The question involves finding the partial derivatives of the function u with respect to x and y, and then combining them in a particular way. The function is given as u = sin-1((x2 + y2) / (x + y)). To find x({partial u} / {partial x}) +

y({partial u} / {partial y}), we need to apply the chain rule to differentiate u with respect to each variable, which can be particularly challenging because u is defined as the inverse sine of a quotient involving both x and y.

First, differentiate u with respect to x, keeping y constant, to get {partial u} / {partial x}. Then, differentiate u with respect to y, keeping x constant, to get {partial u} / {partial y}.

After calculating these derivatives, multiply the first by x and the second by y, and then add the two products to obtain the desired expression.

User JM At Work
by
7.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories