Write the function in factorized form
Given
f ( x ) = ( 2 x + 2 − x ) tan x tan − 1 ( x 2 − x + 1 ) ( 7 x 2 + 3 x + 1 ) 3 f(x)=\frac{(2^x+2^{-x})\tan x\sqrt{\tan^{-1}(x^2-x+1)}}{(7x^2+3x+1)^3} f ( x ) = ( 7 x 2 + 3 x + 1 ) 3 ( 2 x + 2 − x ) tan x tan − 1 ( x 2 − x + 1 ) let
A ( x ) = 2 x + 2 − x , B ( x ) = tan x , C ( x ) = tan − 1 ( x 2 − x + 1 ) , D ( x ) = ( 7 x 2 + 3 x + 1 ) 3 . A(x)=2^x+2^{-x},\quad B(x)=\tan x,\quad C(x)=\sqrt{\tan^{-1}(x^2-x+1)},\quad D(x)=(7x^2+3x+1)^3. A ( x ) = 2 x + 2 − x , B ( x ) = tan x , C ( x ) = tan − 1 ( x 2 − x + 1 ) , D ( x ) = ( 7 x 2 + 3 x + 1 ) 3 . So,
f ( x ) = A ( x ) B ( x ) C ( x ) D ( x ) . f(x)=\frac{A(x)B(x)C(x)}{D(x)}. f ( x ) = D ( x ) A ( x ) B ( x ) C ( x ) . We need f ′ ( 0 ) f'(0) f ′ ( 0 )
Evaluate f ( 0 ) f(0) f ( 0 )
At x = 0 x=0 x = 0
A ( 0 ) = 2 0 + 2 0 = 2 , A(0)=2^0+2^0=2, A ( 0 ) = 2 0 + 2 0 = 2 , B ( 0 ) = tan 0 = 0 , B(0)=\tan 0=0, B ( 0 ) = tan 0 = 0 , C ( 0 ) = tan − 1 ( 0 2 − 0 + 1 ) = tan − 1 ( 1 ) = π 4 = π 2 , C(0)=\sqrt{\tan^{-1}(0^2-0+1)}=\sqrt{\tan^{-1}(1)}=\sqrt{\frac{\pi}{4}}=\frac{\sqrt{\pi}}{2}, C ( 0 ) = tan − 1 ( 0 2 − 0 + 1 ) = tan − 1 ( 1 ) = 4 π = 2 π , D ( 0 ) = ( 1 ) 3 = 1. D(0)=(1)^3=1. D ( 0 ) = ( 1 ) 3 = 1. Hence
f ( 0 ) = 2 ⋅ 0 ⋅ π 2 1 = 0. f(0)=\frac{2\cdot 0\cdot \frac{\sqrt\pi}{2}}{1}=0. f ( 0 ) = 1 2 ⋅ 0 ⋅ 2 π = 0.
Use the fact that only tan x \tan x tan x x = 0 x=0 x = 0
Since B ( 0 ) = 0 B(0)=0 B ( 0 ) = 0 x = 0 x=0 x = 0
Write
f ( x ) = E ( x ) tan x , f(x)=E(x)\tan x, f ( x ) = E ( x ) tan x , where
E ( x ) = ( 2 x + 2 − x ) tan − 1 ( x 2 − x + 1 ) ( 7 x 2 + 3 x + 1 ) 3 . E(x)=\frac{(2^x+2^{-x})\sqrt{\tan^{-1}(x^2-x+1)}}{(7x^2+3x+1)^3}. E ( x ) = ( 7 x 2 + 3 x + 1 ) 3 ( 2 x + 2 − x ) tan − 1 ( x 2 − x + 1 ) . Then
f ′ ( x ) = E ′ ( x ) tan x + E ( x ) sec 2 x . f'(x)=E'(x)\tan x+E(x)\sec^2 x. f ′ ( x ) = E ′ ( x ) tan x + E ( x ) sec 2 x . At x = 0 x=0 x = 0
t a n 0 = 0 , sec 2 0 = 1. tan 0=0,\qquad \sec^2 0=1. t an 0 = 0 , sec 2 0 = 1. Therefore,
f ′ ( 0 ) = E ( 0 ) . f'(0)=E(0). f ′ ( 0 ) = E ( 0 ) . So we only need to compute E ( 0 ) E(0) E ( 0 )
Compute E ( 0 ) E(0) E ( 0 )
E ( 0 ) = ( 2 0 + 2 0 ) tan − 1 ( 1 ) ( 1 ) 3 = 2 π 4 = 2 ⋅ π 2 = π . E(0)=\frac{(2^0+2^0)\sqrt{\tan^{-1}(1)}}{(1)^3}
=2\sqrt{\frac{\pi}{4}}=2\cdot \frac{\sqrt\pi}{2}=\sqrt\pi. E ( 0 ) = ( 1 ) 3 ( 2 0 + 2 0 ) tan − 1 ( 1 ) = 2 4 π = 2 ⋅ 2 π = π . Thus,
f ′ ( 0 ) = π . f'(0)=\sqrt{\pi}. f ′ ( 0 ) = π .
Check options
The correct option is
π \boxed{\sqrt{\pi}} π which is Option B .