Giải câu 2 bài 5: Đạo hàm cấp hai

a) \(y =  \frac{1}{1-x}\)

\(y' =\frac{1'.(1-x)-1.(1-x)'}{(1-x)^{2}}= -\frac{(1-x)'}{(1-x)^{2}} = \frac{1}{(1-x)^{2}}\)

\(y" =  -\frac{[(1-x)^{2}]'}{(1-x)^{4}} = - \frac{2.(-1)(1-x)}{(1-x)^{4}} = \frac{2}{(1-x)^{3}}\).

b) \(y =  \frac{1}{\sqrt{1-x}}\)

\(y' =  -\frac{(\sqrt{1-x})'}{1-x} =  \frac{1}{2\sqrt{1-x}(1-x)}\)

\(y" =  -\frac{1}{2}\frac{[(1-x)\sqrt{1-x}]'}{(1-x)^{3}}\)

\(=-\frac{1}{2}\frac{-\sqrt{1-x}+(1-x)\frac{-1}{2\sqrt{1-x}}}{(1-x)^{3}}\)

\(= \frac{3}{4(1-x)^{2}\sqrt{1-x}}\).

c) \(y = \tan x\)

\(y' =  \frac{1}{cos^{2}x}\)

\(y" =  -\frac{(cos^{2}x)'}{cos^{4}x}  = \frac{2cosx.sinx}{cos^{4}x}=  \frac{2sinx}{cos^{3}x}\).

d) \(y = \cos^2x\)

\(y' = 2cosx.(cosx)' = 2cosx.(-sinx)= - 2sinx.cosx = -sin2x\),

\(y" = -(2x)'.cos2x = -2cos2x\).