Show that d dx tan h -1 x - 1 1- x 2

tanh^ 1x=12ln (1+x1 x ) ⇒ddx (tanh^ 1x) =12 [ (1 x1+x)] #215; [ ((1 x)(1) (1+x)( 1)(1 x)^2 ) ] ⇒ddx (tanh^ 1x) =11 x^2 ...

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First Principle of Differentiation

Show that d...

Question

Show that $\frac{d}{d x} (tan h^{- 1} x) = \frac{1}{1 - x^{2}}$

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\frac{d}{d x} (cos h^{- 1} x) = \frac{1}{\sqrt{x^{2} - 1}}

\frac{d}{d x} (cot h^{- 1} x) = \frac{- 1}{(x^{2} - 1)}

\int \sqrt{(\frac{1 + x}{1 - x})} d x = {sin}^{- 1} x - \sqrt{1 - x^{2}}

{sin}^{- 1} \frac{2 x}{1 + x^{2}}; {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}; {tan}^{- 1} \frac{2 x}{1 - x^{2}} .

2 {tan}^{- 1} x .

\int 2 {tan}^{- 1} x = 2 [x {tan}^{- 1} x - \frac{1}{2} log (1 + x^{2})]

\int \frac{(x^{2} - 1) d x}{(x^{4} + 3 x^{2} + 1) {tan}^{- 1} (x + \frac{1}{x})} = \frac{1}{2} log {tan}^{- 1} (x + \frac{1}{x})

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