Derivative of Arctangent Function

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Theorem

Let $x \in \R$.

Let $\arctan x$ be the arctangent of $x$.


Then:

$\dfrac {\map \d {\arctan x} } {\d x} = \dfrac 1 {1 + x^2}$


Corollary

$\dfrac {\mathrm d \left({\arctan \left({\frac x a}\right) }\right)} {\mathrm d x} = \dfrac a {a^2 + x^2}$


Proof 1

\(\displaystyle y\) \(=\) \(\displaystyle \arctan x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle \tan y\) Definition of Real Arctangent
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d x} {\d y}\) \(=\) \(\displaystyle \sec^2 y\) Derivative of Tangent Function
\(\displaystyle \) \(=\) \(\displaystyle 1 + \tan^2 y\) Difference of Squares of Secant and Tangent
\(\displaystyle \) \(=\) \(\displaystyle 1 + x^2\) Definition of $x$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d y} {\d x}\) \(=\) \(\displaystyle \frac 1 {1 + x^2}\) Derivative of Inverse Function

$\blacksquare$


Proof 2

\(\displaystyle \frac {\map \d {\map \arctan x} } {\d x}\) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \frac {\map \arctan {x + h} - \map \arctan x} h\) Definition of Derivative of Real Function at Point
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \frac {\arctan {x + h} + \map \arctan {-x} } h\) Arctangent Function is Odd
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \frac 1 h \map \arctan {\frac {x + h - x} {1 + x \paren {x + h} } }\) Sum of Arctangents
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \frac 1 h \map \arctan {\frac h {1 + x^2 + h x} }\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \frac 1 h \paren {\frac h {1 + x^2 + h x} - \frac 1 3 \paren {\frac h {1 + x^2 + h x} }^3 + \frac 1 5 \paren {\frac h {1 + x^2 + h x} }^5 + \map \OO {h^7} }\) Definition of Real Arctangent
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \paren {\frac 1 {1 + x^2 + h x} - \frac {h^2} {3 \paren {1 + x^2 + h x}^3} + \frac {h^4} {5 \paren {1 + x^2 + h x}^5} + \map \OO {h^6} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {1 + x^2 + 0 x} - \frac {0^2} {3 \paren {1 + x^2 + 0 x}^3} + \frac {0^4} {5 \paren {1 + x^2 + 0 x}^5}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {1 + x^2}\)

$\blacksquare$


Also defined as

This result can also be reported as:

$\dfrac {\map \d {\arctan x} } {\d x} = \dfrac 1 {x^2 + 1}$


Also see


Sources