Derivative of Arctangent Function/Proof 1
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Theorem
- $\dfrac {\map \d {\arctan x} } {\d x} = \dfrac 1 {1 + x^2}$
Proof
| \(\ds y\) | \(=\) | \(\ds \arctan x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \tan y\) | Definition of Real Arctangent | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {\d x} {\d y}\) | \(=\) | \(\ds \sec^2 y\) | Derivative of Tangent Function | ||||||||||
| \(\ds \) | \(=\) | \(\ds 1 + \tan^2 y\) | Difference of Squares of Secant and Tangent | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1 + x^2\) | Definition of $x$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {\d y} {\d x}\) | \(=\) | \(\ds \frac 1 {1 + x^2}\) | Derivative of Inverse Function |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Differentiation: Inverse Ratios