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

User Hellen
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1 Answer

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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
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