Tính đạo hàm các hàm số sau:
a) $y=\sqrt{x+\sqrt{x+\sqrt{x} } } $                               b) $y=\frac{\sin \sqrt{x+1} }{\cos 5x} $
c) $y=\frac{(x+1)^3}{(x-1)^2} $                                                     d) $y=(3x-1)^2\sqrt{9x^2+3x+1}  $
e) $y=\frac{\sqrt{x^2-1} }{3x} (\frac{1}{x^2}+2 )$                                f) $y=\frac{\sqrt{x^2+1}+x }{\sqrt{x^2+1}-x }+\frac{\sqrt{x^2+1}-x }{\sqrt{x^2+1}+x }   $
a) $y'=\frac{1}{2\sqrt[]{x+\sqrt[]{x+\sqrt[]{x} } } } \left[ {1+\frac{1}{2\sqrt[]{x+\sqrt[]{x} } } } \left ( 1+\frac{1}{2\sqrt[]{x} }  \right ) \right]  $
         $= \frac{1+2\sqrt[]{x} +4\sqrt[]{x} .\sqrt[]{x+\sqrt[]{x} } }{8\sqrt[]{x} .\sqrt[]{x+\sqrt[]{x} }.\sqrt[]{x+\sqrt[]{x+\sqrt[]{x} } }  } $
b) $y'=\frac{\frac{1}{2\sqrt[]{x+1} } \cos \sqrt[]{x+1}.\cos 5x +5\sin \sqrt[]{x+1} \sin 5x   }{\cos^2 5x} $
          $=\frac{\cos 5x. \cos \sqrt[]{x+1}+10\sqrt[]{x+1}.\sin \sqrt[]{x+1}. \sin 5x}{2\sqrt[]{x+1}.\cos^25x } $
c)  $y'=\frac{(x+1)^2(x+5)}{(x-1)^3} $
d)  $y'=\frac{9(3x-1)(18x^2+3x+1)}{2\sqrt[]{9x^2+3x+1} } $
e) Biến đổi   $y=\frac{\sqrt[]{x^2-1}(2x^2+1) }{3x^3}  \Rightarrow  y'=\frac{1}{x^4\sqrt[]{x^3-1} } $

f) Biến đổi về dạng $y=4x^2+2 \Rightarrow  y'=8x$


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