What is the derivative of f ( x ) = tan - 1 ( x ) ?

Question

What is the derivative of             f              (        x        )              =                  tan                  -          1                            (        x        )            ?

Anastasia Macdonald · Accepted Answer

I seem to recall my professor forgetting how to deriving this. This is what I showed him:

y = \arctan x

\tan y = x

\sec^{2} y \frac{d y}{d x} = 1

\frac{d y}{d x} = \frac{1}{\sec^{2} y}

Since

\tan y = \frac{x}{1}

and

\sqrt{1^{2} + x^{2}} = \sqrt{1 + x^{2}}

,

\sec^{2} y = {(\frac{\sqrt{1 + x^{2}}}{1})}^{2} = 1 + x^{2}

\Rightarrow \frac{d y}{d x} = \frac{1}{1 + x^{2}}

I believe this was his original intention:

\frac{d y}{d x} = \frac{1}{\sec^{2} y}

\sec^{2} y = 1 + \tan^{2} y

\tan^{2} y = x \to \sec^{2} y = 1 + x^{2}

\Rightarrow \frac{d y}{d x} = \frac{1}{1 + x^{2}}

What is the derivative of f(x)=tan^(-1)(x)?

Answered question

Answer & Explanation

New Questions in Trigonometry