Question

[....]

1/ Tính đạo hàm:

\(y=\left|x-1\right|\left(x\ne1\right)\) 

bang 2 cach

2/ Dao ham:

\(y=\sqrt{\dfrac{1}{2}+\dfrac{1}{2}\sqrt{\dfrac{1}{2}+\dfrac{1}{2}\sqrt{\dfrac{1}{2}+\dfrac{1}{2}\cos x}}}\left(x\in0;\pi\right)\)

Answer 1

[Akai Haruma]

1. 

Cách 1: Tính bằng công thức

\(\left\{\begin{matrix} y=x-1(x>1)\\ y=1-x(x<1)\end{matrix}\right.\Rightarrow \left\{\begin{matrix} y'=1(x>1)\\ y'=-1(x<1)\end{matrix}\right.\)

Tóm gọn lại: $y'=\frac{|x-1|}{x-1}$

Cách 2: Tính bằng định nghĩa.

\(y'=\lim\limits_{x\to 1}\frac{|x-1|-0}{x-1}=\frac{|x-1|}{x-1}\)

Answer 2

[Akai Haruma]

2. Với $x\in (0;\pi)$ thì:

\(y=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{\cos x+1}{2}}}}=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\cos ^2\frac{x}{2}}}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\cos \frac{x}{2}}}=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\cos ^2\frac{x}{4}}}=\sqrt{\frac{1}{2}+\frac{1}{2}\cos \frac{x}{4}}=\sqrt{\cos ^2\frac{x}{8}}=\cos \frac{x}{8}\)

\(\Rightarrow y'=-\frac{1}{8}\sin \frac{x}{8}\)