Question

find d/dx tan^-1√x is

Answer 1

ANSWER:

$\bf{\frac{d}{dx} ( { \tan}^{ - 1} \sqrt{x} ) = \frac{1}{1 + x } }$

GIVEN:

√x

TO FIND:

$\frac{d}{dx} ( { \tan}^{ - 1} \sqrt{x} )$

FORMULA:

$\frac{d}{dx} ( { \tan}^{ - 1}x ) = \frac{ 1}{1 + {x}^{2} }$

EXPLANATION:

$\frac{d}{dx} ( { \tan}^{ - 1} \sqrt{x} ) = \frac{1}{1 + {( \sqrt{x} )}^{2} }$

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

$\frac{d}{dx} ( { \tan}^{ - 1} \sqrt{x} ) = \frac{1}{1 + x }$

SOME FORMULAE:

$\frac{d}{dx} ( { \sin}^{ - 1}x ) = \frac{1}{ \sqrt{1 - {x}^{2}} } \\ \\ \frac{d}{dx} ( { \cos}^{ - 1}x ) = \frac{ - 1}{ \sqrt{1 - {x}^{2}} } \\ \\ \frac{d}{dx} ( { \cot}^{ - 1}x ) = \frac{ - 1}{1 + {x}^{2} }$

$\frac{d}{dx} (constant) = 0 \\ \\ \frac{d}{dx} x = 1 \\ \\ \frac{d}{dx}( {x}^{2} )= 2x \\ \\ \frac{d}{dx} log(x) = \frac{1}{x} \\ \\ \frac{d}{dx} {x}^{n} = n {x}^{n - 1}$