函数求微分训练集
\(x^{2}y-e^{2x}=\sin{y}\)
\[\begin{align}
x^{2}y-e^{2x}=\sin{y}, \quad 若y=y(x), \quad dy=?
\\ \\
x^{2}y-e^{2x}=\sin{y} \Rightarrow d(x^{2}y)-d(e^{2x})=d(\sin{y})
\\ \\
一步\quad \because y是自变量: \quad d(\sin{y})=(\sin{y})'dy = \cos{y}dy
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二步\quad \because x是自变量: \quad d(e^{2x})=(e^{2x})'dx
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设u=2x,\quad e^{2x}=e^{u} \\ \\
(e^{2x})'=(2x)' \cdot (e^{u})'=2\cdot (x)' \cdot (e^{u})' \\
=2\cdot 1\cdot e^{u}lne=2e^{2x}
\\ \\
\therefore d(e^{2x})=2e^{2x}dx
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三步: \quad \because d(uv)=v \cdot du+u \cdot dv
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\therefore d(x^{2}y)=d(x^2)y+x^{2}dy
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\because d(x^2)=(x^{2})'dx
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\therefore d(x^{2}y)=2xydx+x^{2}dy
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四步:\quad 2xydx+x^{2}dy-2e^{2x}dx=\cos{y}dy
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2xydx-2e^{2x}dx=\cos{y}dy-x^{2}dy
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\therefore dy=\frac{2(xy-e^{2x})}{\cos{y}-x^{2}}dx
\end{align}
\]
求\(\sin31°\)近似值
\[\begin{align}
设f(x)=\sin{x}, \enspace 题意即为:
\\
\enspace 求f(x)在x_{0}=30处附近一点x_{1}=31的近似值
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由公式 f(x_{1})\approx f(x_{0})+f'(x_{0})\cdot (x_{1}-x_{0}) 和\pi =180° 得:
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\therefore f(31°)\approx \sin\frac{\pi}{6}+cos \frac{\pi}{6}\cdot (31°-30°)
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\therefore f(31°) \approx \frac{2}{2}+\frac{\sqrt[]{3}}{2}\cdot \frac{\pi}{180}
\\ \\
f(31°)\approx 0.515
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\therefore \sin{31°} \approx 0.515
\end{align}
\]
\(\sqrt{2x}+\frac{1}{3}\cos{x}\)
\[\begin{align}
求微分: \quad y=\sqrt{2x}+\frac{1}{3}\cos{x}
\\ \\
y'=(\sqrt{2x})'+(\frac{1}{3}\cos{x})'
\Rightarrow
(\sqrt{2x})' \cdot (2x)' + (\frac{1}{3}\cos{x})'
\\ \\
\frac{1}{2\sqrt{2x}} \cdot 2 + \frac{1}{3}(-\sin{x})
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y'=\frac{1}{\sqrt{2x}} - \frac{1}{3}\sin{x}
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据微分公式: \ dy=f'(x)dx 而得:\\
dy=y'dx=(\frac{1}{\sqrt{2x}} - \frac{1}{3}\sin{x})dx
\\ \\
函数之微分即为: \ dy=(\frac{1}{\sqrt{2x}} - \frac{1}{3}\sin{x})dx
\end{align}
\]
