Question

Hola❤


plzz solve it guys..if u are really a genius ✌
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Answer 1

Answer:

Given:-

• $y = { \sin }^{ - 1} ( \frac{2x}{1 + { \times }^{2} } )$

To find:-

• dy/dx

Find:-

$let \: x \: = \tan( \alpha ) = > \alpha = { \tan}^{ - 1} x$

$y = { \sin }^{ - 1}( \frac{2 \tan\alpha }{1 + { \tan }^{2} \alpha } ) \\ y = { \sin }^{ - 1} ( \sin2 \alpha ) \\ y = 2 \alpha = 2 { \tan }^{ - 1} x \\ \frac{dy}{dx} = \frac{d( {2 \tan }^{ - 1} x ) }{dx} = 2 \frac{d( { \tan }^{ - 1} x ) }{ dx} \\ \frac{dy}{dx} = \frac{2}{1 + {x}^{2} }$

Hope its help uh❤

Answer 2

Question :–

$\\ { \bold{find \: \: \: \dfrac{dy}{dx} \: \: \: \: if \: \: y \: = \sin {}^{ - 1} ( \dfrac{2x}{1 + {x}^{2} } ) }}$

ANSWER :–

GIVEN :–

$\\ \: \: { \huge{.}} { \bold{ \: \: A \: \: function \: \: y \: = \sin {}^{ - 1} ( \dfrac{2x}{1 + {x}^{2} } ) }}$

TO FIND :–

$\\ \: \: { \huge{.}}{ \bold{ \: \: \dfrac{dy}{dx} \: \: = ?}}$

SOLUTION :–

• First we have solve the function –

$\\ \implies{ \bold{ \: \: y \: = \sin {}^{ - 1} \left( \dfrac{2x}{1 + {x}^{2} } \right) }}$

• Now put x = tan(t) –

$\\ \implies{ \bold{ \: \: y \: = \sin {}^{ - 1} \left[ \dfrac{2 \tan(t)}{1 + { \tan}^{2} (t)} \right] }}$

$\\ \implies{ \bold{ \: \: y \: = \sin {}^{ - 1} [ \sin(2t)] \: \: \: \: \: \left[ \: \because \: \: \sin(2 \theta) = \dfrac{2 \tan( \theta) }{1 + { \tan}^{2}( \theta) } \right]}}$

$\\ \implies{ \bold{ \: \: y \: = 2t }}$

• Now replace 't' –

$\\ \implies{ \bold{ \: \: y \: = 2 \tan ^{ - 1} (x) }}$

• Now Differentiate with respect to 'x' –

$\\ \implies{ \bold{ \: \: \dfrac{dy}{dx} \: = 2 \left( \dfrac{1}{1 + {x}^{2} } \right) }}$

$\\ \implies \large{ \boxed{ \bold{ \dfrac{dy}{dx} \: = \dfrac{2}{1 + {x}^{2} } }}}$

Used formula :–

$\\ \longrightarrow \: { \bold{ \: \dfrac{d[ { \tan}^{ - 1}(x)]}{dx} \: = \dfrac{1}{1 + {x}^{2} } }}$