Đáp án:
\(y' = \dfrac{-3\sin\left(2\sqrt{\cos3x}\right).\sin3x}{2\sqrt{\cos3x}}\\\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad y = \sin^2\left(\sqrt{\cos3x}\right)\\
\to y' = 2\sin\left(\sqrt{\cos3x}\right).\left[\sin\left(\sqrt{\cos3x}\right)\right]'\\
\to y' = 2\sin\left(\sqrt{\cos3x}\right).\left(\sqrt{\cos3x}\right)'.\cos\left(\sqrt{\cos3x}\right)\\
\to y'= 2\sin\left(\sqrt{\cos3x}\right)\cdot\cos\left(\sqrt{\cos3x}\right)\cdot\dfrac{(\cos3x)'}{2\sqrt{\cos3x}}\\
\to y' = \sin\left(2\sqrt{\cos3x}\right)\cdot\dfrac{(3x)'.(-\sin3x)}{2\sqrt{\cos3x}}\\
\to y' = \dfrac{-3\sin\left(2\sqrt{\cos3x}\right).\sin3x}{2\sqrt{\cos3x}}\\
\end{array}\)
