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

Theorem

Let:

$\map f {x, y} = \map {\tan^2} {x^2 - y^2}$

Then:

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

\(=\)

\(\ds 4 x \map \tan {x^2 - y^2} \map {\sec^2} {x^2 - y^2}\)

\(\ds \map {f_2} {1, 2}\)

\(=\)

\(\ds 8 \tan 3 \sec^2 3\)

Proof

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

\(=\)

\(\ds \dfrac \partial {\partial x} {\map {\tan^2} {x^2 - y^2} }\)

Definition of Partial Derivative

\(\ds \)

\(=\)

\(\ds 2 \map \tan {x^2 - y^2} \map {\sec^2} {x^2 - y^2} \cdot 2 x\)

Derivative of Square of Tangent, Derivative of Square Function, Chain Rule for Derivatives

\(\ds \)

\(=\)

\(\ds 4 x \map \tan {x^2 - y^2} \map {\sec^2} {x^2 - y^2}\)

simplifying

$\Box$

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

\(=\)

\(\ds \dfrac \partial {\partial y} {\map {\tan^2} {x^2 - y^2} }\)

Definition of Partial Derivative

\(\ds \)

\(=\)

\(\ds 2 \map \tan {x^2 - y^2} \map {\sec^2} {x^2 - y^2} \cdot \paren {-2 y}\)

Derivative of Square of Tangent, Derivative of Square Function, Chain Rule for Derivatives

\(\ds \)

\(=\)

\(\ds -4 y \map \tan {x^2 - y^2} \map {\sec^2} {x^2 - y^2}\)

simplifying

\(\ds \leadsto \ \ \)

\(\ds \map {f_2} {1, 2}\)

\(=\)

\(\ds -4 \paren 2 \map \tan {1^2 - 2^2} \map {\sec^2} {1^2 - 2^2}\)

\(\ds \)

\(=\)

\(\ds -8 \map \tan {-3} \map {\sec^2} {-3}\)

\(\ds \)

\(=\)

\(\ds -8 \paren {-\tan 3} \sec^2 3\)

Tangent Function is Odd, Secant Function is Even

\(\ds \)

\(=\)

\(\ds 8 \tan 3 \sec^2 3\)

simplifying

$\blacksquare$

Sources

• 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $3$