Suppose f ( x ) = ( 2 x + 2 - x ) tan x tan - 1 ( x 2 - x + 1 ) ( 7 x 2 + 3 x + 1 ) 3 . Then the value of f ' ( 0 ) is equal to [2024]

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Solution

Q.
Suppose $f (x)$ $= \frac{(2^{x} + 2^{- x}) \tan x \sqrt{\tan^{- 1} (x^{2} - x + 1)}}{{(7 x^{2} + 3 x + 1)}^{3}}$ . $Then the value of f' (0) is equal to$ [2024]

1 $π$

2 $\frac{π}{2}$

3 $0$

4 $\sqrt{π}$

Ans.
(4)

We have, $f (x)$ $= \frac{(2^{x} + 2^{- x}) \tan x \sqrt{\tan^{- 1} (x^{2} - x + 1)}}{{(7 x^{2} + 3 x + 1)}^{3}}$

Let $A (x) = (2^{x} + 2^{- x}), B (x) = \tan x$ and $C (x)$ $= \sqrt{\tan^{- 1} (x^{2} - x + 1)}$

$A' (x) = 2^{x} \log 2 - 2^{- x} \log 2 = (2^{x} - 2^{- x}) \log 2,$

$B' (x) = \sec^{2} x$

and $C' (x)$ $= \frac{1}{2 \sqrt{\tan^{- 1} (x^{2} - x + 1)}} \times \frac{1}{1 + {(x^{2} - x + 1)}^{2}} \times$ $(2 x - 1)$

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

$A (0) = 2, B (0) = 0, and C (0) = \sqrt{\frac{π}{4}} = \frac{\sqrt{π}}{2}$

Also, $A' (0) = 0, B' (0) = 1, and C' (0) = - \frac{1}{2 \sqrt{π}}$

Now, $f' (x)$ $= \frac{[A' (x) B (x) C (x) + A (x) B' (x) C (x) + A (x) B (x) C' (x)] - A (x) B (x) C (x) [3 {(7 x^{2} + 3 x + 1)}^{2} (14 x + 3)]}{{(7 x^{2} + 3 x + 1)}^{6}}$

$f' (0)$ $= \frac{[A' (0) B (0) C (0) + A (0) B' (0) C (0) + A (0) B (0) C' (0)] - A (0) B (0) C (0) [3 {(1)}^{2} (3)]}{{(1)}^{6}}$

$= (0 + 2 \times 1 \times \frac{\sqrt{π}}{2} + 0) - 0 = \sqrt{π}$

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