📝 题目
3．求下列函数在给定点处的导数： （1）$y=\sin x-\cos x$ ，求 $\left.y^{\prime}\right|_{x=\frac{\pi}{6}}$ 和 $\left.y^{\prime}\right|_{x=\frac{\pi}{4}}$ ； （2）$\rho=\theta \sin \theta+\frac{1}{2} \cos \theta$ ，求 $\left.\frac{\mathrm{d} \rho}{\mathrm{d} \theta}\right|_{\theta=\frac{\pi}{4}}$ ； （3）$f(x)=\frac{3}{5-x}+\frac{x^{2}}{5}$ ，求 $f^{\prime}(0)$ 和 $f^{\prime}(2)$ ．
💡 答案与解析
[AI解答]
[AI解答] （1）已知 $ y = \sin x - \cos x $，则 $$ y' = \cos x + \sin x $$ 于是 $$ \left. y' \right|_{x=\frac{\pi}{6}} = \cos\frac{\pi}{6} + \sin\frac{\pi}{6} = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3}+1}{2} $$ $$ \left. y' \right|_{x=\frac{\pi}{4}} = \cos\frac{\pi}{4} + \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} $$
（2）已知 $\rho = \theta \sin\theta + \frac{1}{2}\cos\theta$，则 $$ \frac{\mathrm{d}\rho}{\mathrm{d}\theta} = \sin\theta + \theta\cos\theta - \frac{1}{2}\sin\theta = \frac{1}{2}\sin\theta + \theta\cos\theta $$ 代入 $\theta = \frac{\pi}{4}$： $$ \left.\frac{\mathrm{d}\rho}{\mathrm{d}\theta}\right|_{\theta=\frac{\pi}{4}} = \frac{1}{2}\cdot\frac{\sqrt{2}}{2} + \frac{\pi}{4}\cdot\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4} + \frac{\pi\sqrt{2}}{8} = \frac{\sqrt{2}}{8}(2+\pi) $$
（3）已知 $ f(x) = \frac{3}{5-x} + \frac{x^{2}}{5} $，先求导： $$ f'(x) = 3\cdot \frac{1}{(5-x)^2} + \frac{2x}{5} $$ 于是 $$ f'(0) = 3\cdot\frac{1}{25} + 0 = \frac{3}{25} $$ $$ f'(2) = 3\cdot\frac{1}{(5-2)^2} + \frac{4}{5} = 3\cdot\frac{1}{9} + \frac{4}{5} = \frac{1}{3} + \frac{4}{5} = \frac{5}{15} + \frac{12}{15} = \frac{17}{15} $$
难度：★☆☆☆☆