Question

find all the second partial derivatives.
$z = \arctan(\frac{x + y}{1 - xy})$
$z_{xx} = $
$z_{xy} = $
$z_{yx} = $
$z_{yy} = $

Explanation:

Step1: Simplify the function

Using the trigonometric identity \(\arctan a-\arctan b=\arctan\frac{a - b}{1+ab}\), we know that \(\arctan x+\arctan y=\arctan\frac{x + y}{1-xy}\) (for \(xy< 1\)). So \(z=\arctan x+\arctan y\).

Step2: Find the first - order partial derivatives

• For \(z_x\):

Differentiate \(z\) with respect to \(x\) using the formula \((\arctan u)^\prime=\frac{u^\prime}{1 + u^{2}}\). Since \(z=\arctan x+\arctan y\), then \(z_x=\frac{1}{1+x^{2}}+0=\frac{1}{1+x^{2}}\).

• For \(z_y\):

Differentiate \(z\) with respect to \(y\). Then \(z_y = 0+\frac{1}{1 + y^{2}}=\frac{1}{1+y^{2}}\).

Step3: Find the second - order partial derivatives

• For \(z_{xx}\):

Differentiate \(z_x=\frac{1}{1+x^{2}}\) with respect to \(x\). Using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = 1\), \(v=1 + x^{2}\), \(u^\prime=0\), \(v^\prime = 2x\). Then \(z_{xx}=\frac{0\times(1 + x^{2})-1\times2x}{(1 + x^{2})^{2}}=-\frac{2x}{(1 + x^{2})^{2}}\).

• For \(z_{xy}\):

Differentiate \(z_x=\frac{1}{1+x^{2}}\) with respect to \(y\). Since \(\frac{1}{1+x^{2}}\) is a function of \(x\) only, \(z_{xy}=0\).

• For \(z_{yx}\):

Differentiate \(z_y=\frac{1}{1+y^{2}}\) with respect to \(x\). Since \(\frac{1}{1+y^{2}}\) is a function of \(y\) only, \(z_{yx}=0\).

• For \(z_{yy}\):

Differentiate \(z_y=\frac{1}{1+y^{2}}\) with respect to \(y\). Using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = 1\), \(v=1 + y^{2}\), \(u^\prime=0\), \(v^\prime = 2y\). Then \(z_{yy}=\frac{0\times(1 + y^{2})-1\times2y}{(1 + y^{2})^{2}}=-\frac{2y}{(1 + y^{2})^{2}}\).

Answer:

\(z_{xx}=-\frac{2x}{(1 + x^{2})^{2}}\), \(z_{xy}=0\), \(z_{yx}=0\), \(z_{yy}=-\frac{2y}{(1 + y^{2})^{2}}\)