If f(x)=(logcotxtanx)(logtanxcotx)-1 +tan-1 (4x/4-x2), then f'(2) is equal to

Question

If f(x)=(logcotxtanx)(logtanxcotx)-1 +tan-1 (4x/4-x2), then f'(2) is equal to (A) (1/2) (B) -(1/2) (C) 1 (D) -1

Tardigrade Admin · Accepted Answer

f (x)=(logcot x tan x)(logtan x cot x)-1+tan-1 (4x/4-x2) =(log tan x/log cot x)⋅(log tan x/log cot x)+tan-1((4x/4-x2)) =((log tan x2)/(-log tan x2))+tan-1((4x/4-x2)) =1+tan-1((4x/4-x2)) ∴ f '(x)=(1/1+((4x)4-x2)2)⋅((4-x2)4-4x(-2x)/(4-x2)2) =(16-4x2+8x2/(4-x2)2+16x2) =(4(4+x2)/(4-x2)2+(4x)2) Hence, f '(2)=(4(4+4)/0+(8)2) =(30/64)=(1/2)

Q. If $f (x) = (l o g_{co t x} t an x) (l o g_{t an x} co t x)^{- 1}$ $+ t a n^{- 1} \frac{4 x}{4 - x ^{2}}$ , then $f^{'} (2)$ is equal to

$\frac{1}{2}$

$- \frac{1}{2}$

$1$

$- 1$

Q. If f(x)=(logcotx​tanx)(logtanx​cotx)−1 +tan−14−x24x​, then f′(2) is equal to

21​

−21​

1

−1

Q. If $f (x) = (l o g_{co t x} t an x) (l o g_{t an x} co t x)^{- 1}$ $+ t a n^{- 1} \frac{4 x}{4 - x ^{2}}$ , then $f^{'} (2)$ is equal to

$\frac{1}{2}$

$- \frac{1}{2}$

$1$

$- 1$