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Question: If we have a function as \[u = {\sin ^{ - 1}}\left[ {\dfrac{{\left( {{x^2} + {y^2}} \right)}}{{\left...

If we have a function as u=sin⁡−1[(x2+y2)(x+y)]u = {\sin ^{ - 1}}\left[ {\dfrac{{\left( {{x^2} + {y^2}} \right)}}{{\left( {x + y} \right)}}} \right] , then x∂u∂x+y∂u∂yx\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} is equal to
1. sin⁡u\sin u
2. tan⁡u\tan u
3. cos⁡u\cos u
4. cot⁡u\cot u

Explanation

Solution

Here in this question we are given a function u having variables x and y . We will operate sine on both the sides so as to simplify the function. We will use the concept of homogeneous functions here. We need to find the derivative of u partially with respect to x and partially with respect to y. we have to find the value for the expression x∂u∂x+y∂u∂yx\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} . We will be using Euler’s formula to find the value of the required expression.

Complete step-by-step solution:
Here in this question we are given a function u=sin⁡−1[(x2+y2)(x+y)]u = {\sin ^{ - 1}}\left[ {\dfrac{{\left( {{x^2} + {y^2}} \right)}}{{\left( {x + y} \right)}}} \right] and we have to find the value of the expression x∂u∂x+y∂u∂yx\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} .
Homogeneous function: A homogeneous function is one with multiplicative scaling behavior that is if all its arguments are multiplied by a factor, then its value is multiplied by some power of this factor.
We are give u=sin⁡−1[(x2+y2)(x+y)]u = {\sin ^{ - 1}}\left[ {\dfrac{{\left( {{x^2} + {y^2}} \right)}}{{\left( {x + y} \right)}}} \right]
Therefore sin⁡u=[(x2+y2)(x+y)]\sin u = \left[ {\dfrac{{\left( {{x^2} + {y^2}} \right)}}{{\left( {x + y} \right)}}} \right]
Taking x2{x^2}common from the numerator and xxfrom the denominator we get ,
sin⁡u=[x2(1+y2x2)x(1+yx)]\sin u = \left[ {\dfrac{{{x^2}\left( {1 + \dfrac{{{y^2}}}{{{x^2}}}} \right)}}{{x\left( {1 + \dfrac{y}{x}} \right)}}} \right]
=xf(yx)= xf\left( {\dfrac{y}{x}} \right)
This is a homogeneous function of degree 11.
So by Euler’s theorem
x∂z∂x+y∂z∂y=nzx\dfrac{{\partial z}}{{\partial x}} + y\dfrac{{\partial z}}{{\partial y}} = nz where n =degree of homogeneous equation.
Let z=sin⁡uz = \sin u
x∂∂xsin⁡u+y∂∂ysin⁡u=sin⁡ux\dfrac{\partial }{{\partial x}}\sin u + y\dfrac{\partial }{{\partial y}}\sin u = \sin u
Hence on solving the derivatives in the above equation we get ,
xcos⁡u∂u∂x+ycos⁡u∂u∂y=sin⁡ux\cos u\dfrac{{\partial u}}{{\partial x}} + y\cos u\dfrac{{\partial u}}{{\partial y}} = \sin u
Taking cos⁡u\cos u common from LHS of the expression we get ,
x∂u∂x+y∂u∂y=sin⁡ucos⁡ux\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = \dfrac{{\sin u}}{{\cos u}}
On simplification of the expression we get ,
x∂u∂x+y∂u∂y=tan⁡ux\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = \tan u
Hence we get the value of the required expression.
Hence option (2) is the answer.

Note: In order to solve such types of questions one must have a grip on the concept of derivatives very much as we have to find the derivative for various functions in such questions. Keep in mind that here we have computed partial derivatives. We must do the calculations carefully and should recheck at each and every step in order to get the exact solution.