Partial Derivatives of x tan^-1 (x^2 + y)

Theorem

Let:

$\map f {x, y} = x \map \arctan {x^2 + y}$

Then:

\(\ds \map {f_1} {1, 0}\)

\(=\)

\(\ds \dfrac \pi 4 + 1\)

\(\ds \map {f_2} {x, y}\)

\(=\)

\(\ds \dfrac x {1 + \paren {x^2 + y}^2}\)

Proof

\(\ds \map {f_1} {x, y}\)

\(=\)

\(\ds \map {\dfrac \partial {\partial x} } {x \map \arctan {x^2 + y} }\)

Definition of Partial Derivative

\(\ds \)

\(=\)

\(\ds \map \arctan {x^2 + y} \map {\dfrac \partial {\partial x} } x + x \map {\dfrac \partial {\partial x} } {\map \arctan {x^2 + y} }\)

Product Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \map \arctan {x^2 + y} + x \cdot \dfrac 1 {1 + \paren {x^2 + y}^2} \cdot 2 x\)

Product Rule for Derivatives, Derivative of Arctangent Function, Chain Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \map \arctan {x^2 + y} + \dfrac {2 x^2} {1 + \paren {x^2 + y}^2}\)

simplifying

\(\ds \leadsto \ \ \)

\(\ds \map {f_1} {1, 0}\)

\(=\)

\(\ds \map \arctan {1^2 + 0} + \dfrac {2 \times 1^2} {1 + \paren {1^2 + 0}^2}\)

substituting $x = 1$ and $y = 0$

\(\ds \)

\(=\)

\(\ds \map \arctan 1 + \dfrac 2 2\)

simplifying

\(\ds \)

\(=\)

\(\ds \dfrac \pi 4 + 1\)

Tangent of $45 \degrees$

$\Box$

\(\ds \map {f_2} {x, y}\)

\(=\)

\(\ds \map {\dfrac \partial {\partial y} } {x \map \arctan {x^2 + y} }\)

Definition of Partial Derivative

\(\ds \)

\(=\)

\(\ds x \dfrac 1 {1 + \paren {x^2 + y}^2} \cdot 1\)

Derivative of Identity Function, Derivative of Arctangent Function, Chain Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \dfrac x {1 + \paren {x^2 + y}^2}\)

simplifying

$\blacksquare$

Sources

• 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $6$