Đáp án:
\(d)y' = - \dfrac{{\sin x}}{{4\sqrt {1 + {{\cos }^2}\dfrac{x}{2}} }}\)
Giải thích các bước giải:
\(\begin{array}{l}
b)y' = \dfrac{{ - x.\sin x - \cos x}}{{{x^2}}} + \dfrac{{\sin x - x.\cos x}}{{{{\sin }^2}x}}\\
= \dfrac{{ - x.{{\sin }^3}x - {{\sin }^2}x.\cos x + {x^2}.\sin x - {x^3}.\cos x}}{{{x^2}.{{\sin }^2}x}}\\
d)y' = \left( {2.\dfrac{1}{2}.\cos \dfrac{x}{2}\left( { - \sin \dfrac{x}{2}} \right)} \right).\dfrac{1}{{2\sqrt {1 + {{\cos }^2}\dfrac{x}{2}} }}\\
= \dfrac{{ - \cos \dfrac{x}{2}.\sin \dfrac{x}{2}}}{{2\sqrt {1 + {{\cos }^2}\dfrac{x}{2}} }} = \dfrac{{ - \dfrac{{\sin x}}{2}}}{{2\sqrt {1 + {{\cos }^2}\dfrac{x}{2}} }}\\
= - \dfrac{{\sin x}}{{4\sqrt {1 + {{\cos }^2}\dfrac{x}{2}} }}
\end{array}\)
