Partial Derivatives of x tan^-1 (x^2 + y)
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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 $6$