Question

7.(7 pts.) the equation of the tangent line to $y = \sin(x^2) + 1$ at $(x, y) = (\sqrt{\pi/2}, 2)$ is
(a) $y = (2\sqrt{\pi/2}) x + 2$ (b) $x = 2$
(c) $y = 2$ (d) $y = x + 2 - \sqrt{\pi/2}$
(e) $y = \pi(x - \sqrt{\pi/2}) + 2$.

Explanation:

Step1: 求函数的导数

函数\(y = \sin(x^2)+1\)，根据复合函数求导法则，令\(u = x^2\)，则\(y=\sin u + 1\)，先对\(y\)关于\(u\)求导，\(y^\prime_{u}=\cos u\)，再对\(u\)关于\(x\)求导，\(u^\prime_{x}=2x\)，根据复合函数求导公式\(y^\prime_{x}=y^\prime_{u}\cdot u^\prime_{x}\)，可得\(y^\prime = 2x\cos(x^2)\)。

Step2: 计算切线的斜率

将\(x = \sqrt{\frac{\pi}{2}}\)代入到导数\(y^\prime\)中，计算斜率\(k\)。此时\(x^2=\frac{\pi}{2}\)，则\(\cos(x^2)=\cos(\frac{\pi}{2}) = 0\)？不对，重新计算：\(x=\sqrt{\frac{\pi}{2}}\)，则\(x^2=\frac{\pi}{2}\)，\(\cos(x^2)=\cos(\frac{\pi}{2}) = 0\)？不，等一下，\(y^\prime=2x\cos(x^2)\)，当\(x = \sqrt{\frac{\pi}{2}}\)时，\(x^2=\frac{\pi}{2}\)，\(\cos(\frac{\pi}{2}) = 0\)？不对，\(\cos(\frac{\pi}{2}) = 0\)吗？是的，\(\cos(\frac{\pi}{2}) = 0\)，那这样斜率\(k = 2\times\sqrt{\frac{\pi}{2}}\times\cos(\frac{\pi}{2})=0\)？不对，可能我哪里错了。哦，不对，\(y=\sin(x^2)+1\)，求导应该是\(y^\prime = \cos(x^2)\times2x\)，当\(x = \sqrt{\frac{\pi}{2}}\)时，\(x^2=\frac{\pi}{2}\)，\(\cos(\frac{\pi}{2}) = 0\)？不，\(\cos(\frac{\pi}{2}) = 0\)，所以斜率\(k = 2\times\sqrt{\frac{\pi}{2}}\times0 = 0\)？不对，可能我计算错误。或者，重新看，\(x=\sqrt{\frac{\pi}{2}}\)，则\(x^2=\frac{\pi}{2}\)，\(\cos(x^2)=\cos(\frac{\pi}{2}) = 0\)，所以导数为\(0\)，即切线斜率为\(0\)。

Step3: 利用点斜式求切线方程

点斜式方程为\(y - y_0=k(x - x_0)\)，其中\((x_0,y_0)=(\sqrt{\frac{\pi}{2}},2)\)，\(k = 0\)，代入可得\(y - 2=0\times(x - \sqrt{\frac{\pi}{2}})\)，即\(y = 2\)。

Answer:

(c) \(y = 2\)