m.n ơi giúp e vs ạ e cảm ơn nhìu ^-^ Tính đạo hàm a)căn bậc hai(1 +2*x-x^2) b) √(x ²+1) – √ (1-x²) c)cos(2x)tanx

a) ($\sqrt[]{1+2^x-x^2}$ )’ 

= $\dfrac{(1+2^x-x^2)’}{2\sqrt[]{1+2^x-x^2}}$ 

= $\dfrac{x.2^{x-1}-2x}{2\sqrt[]{1+2^x-x^2}}$ 

.

b) ($\sqrt[]{x^2+1}$ – $\sqrt[]{1-x^2}$ )’

= $\dfrac{(x^2+1)’}{2\sqrt[]{x^2+1}}$ – $\dfrac{(1-x^2)’}{2\sqrt[]{1-x^2}}$ 

= $\dfrac{2x}{2\sqrt[]{x^2+1}}$ – $\dfrac{-2x}{2\sqrt[]{1-x^2}}$ 

.

c) $\text{ (cos 2x . tan x)’}$

= $\text{ (cos 2x)’ . tan x+cos 2x . (tan x)’}$

= $\text{ – sin 2x.(2x)’ . tan x+cos 2x . $\dfrac{1}{cos^2x}$}$ 

= $\text{ – 2sin 2x. tan x+ $\dfrac{cos 2x}{cos^2x}$}$