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Question Number 106847 by mathocean1 last updated on 07/Aug/20

G i v e n f (x) = \frac{3 \sqrt{3}}{s i n x} + \frac{1}{c o s x}

s h o w t h a t f^{'} (x) = c o s x \frac{({t a n}^{3} x - 3 \sqrt{3})}{{s i n}^{2} x}

Answered by bemath last updated on 07/Aug/20

^{@ bemath @}

f (x) = 3 \sqrt{3} \csc x + \sec x

f^{'} (x) = - 3 \sqrt{3} \cot x \csc x + \sec x \tan x

f^{'} (x) = \frac{\sin x}{\cos^{2} x} - \frac{3 \sqrt{3} \cos x}{\sin^{2} x}

= \frac{\sin^{3} x - 3 \sqrt{3} \cos^{3} x}{\cos^{2} x \sin^{2} x}

= \frac{\tan^{3} x - 3 \sqrt{3}}{\sin^{2} x (\frac{1}{\cos x})} = \frac{\cos x (\tan^{3} x - 3 \sqrt{3})}{\sin^{2} x}

Commented by mathocean1 last updated on 07/Aug/20

P l e a s e s i r c a n y o u d e t a i l t h e l a s t

l i n e ? I^{'} m n o t u n d e r s t a n d i n g h o w

{t a n}^{3} a p p e a r e d \dots

Commented by bemath last updated on 07/Aug/20

from \frac{\sin^{3} x - 3 \sqrt{3} \cos^{3} x}{\cos^{2} x \sin^{2} x} : \frac{\cos^{3} x}{\cos^{3} x}

Answered by Dwaipayan Shikari last updated on 07/Aug/20

f (x) = \frac{3 \sqrt{3}}{s i n x} + \frac{1}{c o s x}

f^{'} (x) = \frac{- 3 \sqrt{3} c o s x}{{s i n}^{2} x} + s e c x t a n x = \frac{- 3 \sqrt{3} c o s x}{{s i n}^{2} x} + \frac{s i n x}{{c o s}^{2} x}

= \frac{- 3 \sqrt{3} {c o s}^{3} x + {s i n}^{3} x}{{s i n}^{2} {x c o s}^{2} x} = \frac{- 3 \sqrt{3} + \frac{{s i n}^{3} x}{{c o s}^{3} x}}{{s i n}^{2} x \frac{{c o s}^{2} x}{{c o s}^{3} x}} = c o s x (\frac{{t a n}^{3} x - 3 \sqrt{3}}{{s i n}^{2} x})

Commented by bemath last updated on 07/Aug/20

typo sir . it \tan^{3} x

Answered by 1549442205PVT last updated on 07/Aug/20

f (x) = \frac{3 \sqrt{3}}{s i n x} + \frac{1}{c o s x}

\Rightarrow f^{'} (x) = - \frac{3 \sqrt{3}}{\sin^{2} x} \times cosx + \frac{- 1}{\cos^{2} x} \times (- sinx)

(since {(\frac{1}{u})}^{'} = \frac{- 1}{u^{2}} \times u^{'})

= \frac{- 3 \sqrt{3} cosx}{\sin^{2} x} + \frac{sinx}{\cos^{2} x} = \frac{- 3 \sqrt{3} cosx}{\sin^{2} x} + \frac{\sin^{3} x}{\sin^{2} {xcos}^{2} x}

= \frac{- 3 \sqrt{3} cosx}{\sin^{2} x} + \frac{(\frac{\sin^{3} x}{\cos^{3} x}) cosx}{\sin^{2} x}

= \frac{- 3 \sqrt{3} cosx}{\sin^{2} x} + \frac{\tan^{3} x . cosx}{\sin^{2} x}

= cosx (\frac{\tan^{3} x - 3 \sqrt{3}}{\sin^{2} x}) (q . e . d)

Commented by mathocean1 last updated on 08/Aug/20

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