For the function z = tan^-1 \(\frac{y}{x}\) consider the following statements :

\({S_1} \equiv x\frac{{\partial z}}{{\partial x}} + y\frac{{\partial z}}{{\partial y}} = 0\)

\({S_2} \equiv {x^2}\frac{{{\partial ^2}z}}{{\partial {x^2}}} + 2xy\frac{{{\partial ^2}z}}{{\partial x\partial y}} + {y^2}\frac{{{\partial ^2}z}}{{\partial {y^2}}} = 0\)

\({S_3} \equiv \frac{{{\partial ^2}z}}{{\partial {x^2}}} + \frac{{{\partial ^2}z}}{{\partial {y^2}}} = 0\)

Which of the following is true?

Concept:

In partial differentiation, all variables are considered as a constant except the independent derivative variable i.e. if f(x, y) is a function, then its partial derivative with respect to x is calculated by keeping y as constant and with respect to y is calculated by keeping x as constant.

Formula Used:

\(\frac{d}{dx} tan^{-1}x = \frac{1}{(1 \ + \ x^2)}\)

Calculation:

We have, 

⇒ z = tan^-1\(\frac{y}{x}\)

Differentiate both sides with respect to y, we get

⇒ \(\frac{∂z}{∂y} = \frac{1}{1 \ + \ \frac{y^2}{x^2}} \times \frac{1}{x}\)

⇒ \(\frac{∂z}{∂y} = \frac{x}{x^2 \ + \ y^2}\)    -------(1)

Differentiating again with respect to x, we get

⇒ \(\frac{∂^2z}{∂x∂y} = \frac{-2x^2}{(x^2 \ + \ y^2)^2} \ + \ \frac{1}{x^2 \ + \ y^2}\)    - ---(2)

Differentiating again with respect to y, we get

⇒ \(\frac{∂^2z}{∂y^2} = x \left( \frac{d}{dx} \frac{1}{x^2 \ + \ y^2}\right) \)

⇒ \(\frac{∂^2z}{∂y^2} = \frac{-x}{(x^2 \ + \ y^2)^2} \times 2y\)

⇒ \(\frac{∂^2z}{∂y^2} = \frac{-2xy}{(x^2 \ + \ y^2)^2}\)    -----(3)

Differentiate both sides with respect to x, we get

⇒ \(\frac{∂z}{∂x} = \frac{1}{1 \ + \ \frac{y^2}{x^2}} \times \frac{d}{dx} \frac{y}{x}\)

⇒ \(\frac{∂z}{∂x} = \frac{x^2}{x^2 \ + \ y^2} \times \frac{-1}{x^2} \times y\)

⇒ \(\frac{∂z}{∂x} = \frac{-y}{x^2 \ + \ y^2}\)     ----- (4)

Differentiating again with respect to x, we get

⇒ \(\frac{∂^2z}{∂x^2} = \frac{2xy}{(x^2 \ + \ y^2)^2}\)    ------(5)

Statement 1: \( x\frac{{\partial z}}{{\partial x}} + y\frac{{\partial z}}{{\partial y}} = 0\)

⇒ \(x \times \frac{-y}{x^2 \ + \ y^2} \ + \ y \times \frac{x}{x^2 \ + \ y^2}\)

⇒ \(\frac{xy}{x^2 \ + \ y^2} \ - \ \frac{xy}{x^2 \ + \ y^2}\)

⇒ 0

Statement 1 is true.

Statement 2: \( {x^2}\frac{{{\partial ^2}z}}{{\partial {x^2}}} + 2xy\frac{{{\partial ^2}z}}{{\partial x\partial y}} + {y^2}\frac{{{\partial ^2}z}}{{\partial {y^2}}} = 0\)

⇒ \(\frac{2x^3 y}{(x^2 \ + \ y^2)^2} \ + \ \frac{-4x^3y}{(x^2 \ + \ y^2) ^2} \ + \frac{2xy}{(x^2 \ + \ y^2)} \ +\ \frac{-2xy^3}{(x^2 \ + \ y^2)^2}\)

⇒ \(\frac{ 2x^3 y \ - \ 4x^3y \ + \ 2x^3y \ + \ 2xy^3 - 2xy^3}{(x^2 \ + \ y^2)^2}\)

⇒ \(\frac{4x^3y \ -\ 4x^3y \ + \ 2xy^3 \ - \ 2xy^3}{(x^2 \ + \ y^2)^2} \)

⇒ 0

Statement 2 is true.

Statement 3: \( \frac{{{\partial ^2}z}}{{\partial {x^2}}} + \frac{{{\partial ^2}z}}{{\partial {y^2}}} = 0\)

⇒ \(\frac{2xy}{(x^2 \ + \ y^2)^2} \ + \ \frac{-2xy}{(x^2\ + \ y^2)^2}\)

⇒ 0

Statement 3 is true.

∴ For the function z = tan-1 \(\frac{y}{x}\) all the statements are true.