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+ $\dfrac{cos 2x}{cos^2x}$}$
