Giải bài 5.67 - 5.68 trang 209 - SBT Đại số và Giải tích lớp 11

5.67
$\begin{aligned} & y=\sin \left( {{\cos }^{2}}x \right).\cos \left( {{\sin }^{2}}x \right) \\ & \Rightarrow y'=\left( {{\cos }^{2}}x \right)'\cos \left( {{\cos }^{2}}x \right).\cos \left( {{\sin }^{2}}x \right)+\sin \left( {{\cos }^{2}}x \right).\left[ -\left( {{\sin }^{2}}x \right)'\sin \left( {{\sin }^{2}}x \right) \right] \\ & =-2\sin x\cos x.\cos \left( {{\cos }^{2}}x \right)\cos \left( {{\sin }^{2}}x \right)+\sin \left( {{\cos }^{2}}x \right).\left( -2\sin x\cos x \right)\sin \left( {{\sin }^{2}}x \right) \\ & =-\sin 2x\left[ \cos \left( {{\cos }^{2}}x \right)\cos \left( {{\sin }^{2}}x \right)+\sin \left( {{\cos }^{2}}x \right)\sin \left( {{\sin }^{2}}x \right) \right] \\ & =-\sin 2x\cos \left( {{\cos }^{2}}x-{{\sin }^{2}}x \right) \\ & =-\sin 2x\cos \left( \cos 2x \right) \\ \end{aligned}$
5.68
$\begin{aligned} & y=\dfrac{\sin x-x\cos x}{\cos x+x\sin x} \\ & \Rightarrow y'=\dfrac{\left( \cos x-\cos x+x\sin x \right)\left( \cos x+x\sin x \right)-\left( -\sin x+\sin x+x\cos x \right)\left( \sin x-x\cos x \right)}{{{\left( \cos x+x\sin x \right)}^{2}}} \\ & =\dfrac{x\sin x\left( \cos x+x\sin x \right)-x\cos x\left( \sin x-x\cos x \right)}{{{\left( \cos x+x\sin x \right)}^{2}}} \\ & =\dfrac{{{x}^{2}}{{\sin }^{2}}x+{{x}^{2}}{{\cos }^{2}}x}{{{\left( \cos x+x\sin x \right)}^{2}}}=\dfrac{{{x}^{2}}}{{{\left( \cos x+x\sin x \right)}^{2}}} \\ \end{aligned}$