Cho hàm số \(y = \sin {x^{\sqrt {\cos x} }}\) ta có:
A.\(y'({\pi \over 4}) = {e^{{{ - 1} \over {2\root 4 \of 2 }}\ln 2}}({1 \over {\root 4 \of 2 }} + {1 \over {4\root 4 \of 2 }}\ln 2)\)
B.\(y'({\pi \over 4}) = {e^{{{ - 1} \over {2\sqrt 2 }}\ln 2}}({1 \over {\sqrt 2 }} + {1 \over {2\sqrt 2 }}\ln 2)\)
C.\(y'({\pi \over 4}) = {e^{{1 \over {2\root 4 \of 2 }}\ln 2}}({1 \over {\root 4 \of 2 }} + {1 \over {4\root 4 \of 2 }}\ln 2)\)
D.\(y'({\pi \over 4}) = {e^{{1 \over {2\sqrt 2 }}\ln 2}}({1 \over {\sqrt 2 }} - {1 \over {2\sqrt 2 }}\ln 2)\)
A.\(y'({\pi \over 4}) = {e^{{{ - 1} \over {2\root 4 \of 2 }}\ln 2}}({1 \over {\root 4 \of 2 }} + {1 \over {4\root 4 \of 2 }}\ln 2)\)
B.\(y'({\pi \over 4}) = {e^{{{ - 1} \over {2\sqrt 2 }}\ln 2}}({1 \over {\sqrt 2 }} + {1 \over {2\sqrt 2 }}\ln 2)\)
C.\(y'({\pi \over 4}) = {e^{{1 \over {2\root 4 \of 2 }}\ln 2}}({1 \over {\root 4 \of 2 }} + {1 \over {4\root 4 \of 2 }}\ln 2)\)
D.\(y'({\pi \over 4}) = {e^{{1 \over {2\sqrt 2 }}\ln 2}}({1 \over {\sqrt 2 }} - {1 \over {2\sqrt 2 }}\ln 2)\)
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Hóa Học lớp 12
