d-dx-tan-1-4x-1-4x-2-or-d-dx-tan-1-2tan-where-2x-sin-which-comes-later-if-done-considering-2x-sin-please-help-

Question Number 67960 by aseer imad last updated on 02/Sep/19

\frac{d}{d x} [{t a n}^{- 1} \frac{4 x}{\sqrt{1 - 4 x^{2}}}]

o r

\frac{d}{d x} {t a n}^{- 1} (2 t a n θ) [w h e r e 2 x = s i n θ]

w h i c h c o m e s l a t e r i f d o n e c o n s i d e r i n g

2 x = s i n θ

p l e a s e h e l p

Commented by Prithwish sen last updated on 02/Sep/19

Let y = \tan^{- 1} \frac{4 x}{\sqrt{1 - {4 x}^{2}}}

Now considering 2 x = \sin θ

diff . w . r . t θ we get \frac{dx}{d θ} = \frac{1}{2} Cos θ = \frac{1}{2} \sqrt{1 - {4 x}^{2}}

Now the original expression turns out

y = \tan^{- 1} (2 \tan θ) again diff . w . r . t θ

\frac{dy}{d θ} = \frac{1}{1 + {4 \tan}^{2} θ} . {2 \sec}^{2} θ = \frac{2}{1 + {12 x}^{2}}

∴ \frac{dy}{dx} = \frac{dy}{d θ} . \frac{d θ}{dx} = \frac{4}{(1 + 12 x^{2}) \sqrt{1 - 4 x^{2}}}

It can be done in direct method

\frac{dy}{dx} = \frac{1}{1 + {(\frac{4 x}{\sqrt{1 - {4 x}^{2}}})}^{2}} . \frac{4 \sqrt{1 - {4 x}^{2}} + 4 x . \frac{8 x}{2 \sqrt{1 - {4 x}^{2}}}}{1 - {4 x}^{2}}

= \frac{1 - {4 x}^{2}}{1 - {4 x}^{2} + {16 x}^{2}} . \frac{4 (1 - {4 x}^{2}) + {16 x}^{2}}{(1 - {4 x}^{2}) \sqrt{1 - {4 x}^{2}}}

= \frac{4}{(1 + 12 x^{2}) \sqrt{1 - 4 x^{2}}} please check .

Commented by MJS last updated on 02/Sep/19

\frac{d}{d x} [\arctan u (x)] = \frac{u^{'} (x)}{{(u (x))}^{2} + 1}

u (x) = \frac{4 x}{\sqrt{1 - 4 x^{2}}}

u^{'} (x) = \frac{4}{\sqrt{{(1 - 4 x^{2})}^{3}}}

{(u (x))}^{2} + 1 = \frac{12 x^{2} + 1}{1 - 4 x^{2}}

\frac{u^{'} (x)}{{(u (x))}^{2} + 1} = \frac{4}{(12 x^{2} + 1) \sqrt{1 - 4 x^{2}}}

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