题目
Problem
5.
y = 4 + ln x x > 1 2 \begin{align*}
y = \sqrt{4 + \ln x} \qquad x > \frac{1}{2}
\end{align*} y = 4 + ln x x > 2 1 (a) Show that
d 2 y d x 2 = − 9 + 2 ln x 4 x 2 ( 4 + ln x ) 3 2 \begin{align*}
&\,\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\\[4mm]
=&\,-\frac{9+2\ln x}{4x^2(4+\ln x)^{\frac32}}
\end{align*} = d x 2 d 2 y − 4 x 2 ( 4 + ln x ) 2 3 9 + 2 ln x (5)
(b) Hence, or otherwise, determine the Taylor series expansion about x = 1 x = 1 x = 1 y y y ( x − 1 ) (x - 1) ( x − 1 ) ( x − 1 ) 2 (x - 1)^2 ( x − 1 ) 2
(3)
题目中文翻译
5.
y = 4 + ln x x > 1 2 . \begin{align*}
y=\sqrt{4+\ln x}\qquad x>\frac12.
\end{align*} y = 4 + ln x x > 2 1 . (a) 证明
d 2 y d x 2 = − 9 + 2 ln x 4 x 2 ( 4 + ln x ) 3 2 . \begin{align*}
&\,\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\\[4mm]
=&\,-\frac{9+2\ln x}{4x^2(4+\ln x)^{\frac32}}.
\end{align*} = d x 2 d 2 y − 4 x 2 ( 4 + ln x ) 2 3 9 + 2 ln x . (5)
(b) 由此或用其他方法，求 y y y x = 1 x=1 x = 1 ( x − 1 ) (x-1) ( x − 1 ) ( x − 1 ) 2 (x-1)^2 ( x − 1 ) 2
(3)
解答
(a)
解法一
思路
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把
y = ( 4 + ln x ) 1 2 \begin{align*}
y=(4+\ln x)^{\frac12}
\end{align*} y = ( 4 + ln x ) 2 1 直接求导两次。第二次求导时可以把一阶导数写成
1 2 x − 1 ( 4 + ln x ) − 1 2 , \begin{align*}
\frac12x^{-1}(4+\ln x)^{-\frac12},
\end{align*} 2 1 x − 1 ( 4 + ln x ) − 2 1 , 这样用 product rule 比较清楚。
答题过程
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Given
y = ( 4 + ln x ) 1 2 . \begin{align*}
y=(4+\ln x)^{\frac12}.
\end{align*} y = ( 4 + ln x ) 2 1 . Differentiate:
d y d x = 1 2 ( 4 + ln x ) − 1 2 ⋅ 1 x = 1 2 x ( 4 + ln x ) 1 2 . \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,\frac12(4+\ln x)^{-\frac12}\cdot\frac1x\\[2mm]
=&\,\frac{1}{2x(4+\ln x)^{\frac12}}.
\end{align*} d x d y = = 2 1 ( 4 + ln x ) − 2 1 ⋅ x 1 2 x ( 4 + ln x ) 2 1 1 . Now write
d y d x = 1 2 x − 1 ( 4 + ln x ) − 1 2 . \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,\frac12x^{-1}(4+\ln x)^{-\frac12}.
\end{align*} d x d y = 2 1 x − 1 ( 4 + ln x ) − 2 1 . Differentiate again:
d 2 y d x 2 = 1 2 [ − x − 2 ( 4 + ln x ) − 1 2 + x − 1 ( − 1 2 ) ( 4 + ln x ) − 3 2 ⋅ 1 x ] = − 1 2 x 2 ( 4 + ln x ) 1 2 − 1 4 x 2 ( 4 + ln x ) 3 2 . \begin{align*}
\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
=&\,\frac12\bigg[-x^{-2}(4+\ln x)^{-\frac12}\\[4mm]
&\,\hspace{2pt}+x^{-1}\Big(-\frac12\Big)
(4+\ln x)^{-\frac32}\cdot\frac1x\bigg]\\[4mm]
=&\,-\frac{1}{2x^2(4+\ln x)^{\frac12}}\\[4mm]
&\,\hspace{2pt}-\frac{1}{4x^2(4+\ln x)^{\frac32}}.
\end{align*} d x 2 d 2 y = = 2 1 [ − x − 2 ( 4 + ln x ) − 2 1 + x − 1 ( − 2 1 ) ( 4 + ln x ) − 2 3 ⋅ x 1 ] − 2 x 2 ( 4 + ln x ) 2 1 1 − 4 x 2 ( 4 + ln x ) 2 3 1 . Use common denominator
4 x 2 ( 4 + ln x ) 3 2 . \begin{align*}
4x^2(4+\ln x)^{\frac32}.
\end{align*} 4 x 2 ( 4 + ln x ) 2 3 . Then
d 2 y d x 2 = − 2 ( 4 + ln x ) 4 x 2 ( 4 + ln x ) 3 2 − 1 4 x 2 ( 4 + ln x ) 3 2 = − 8 + 2 ln x + 1 4 x 2 ( 4 + ln x ) 3 2 = − 9 + 2 ln x 4 x 2 ( 4 + ln x ) 3 2 . \begin{align*}
\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
=&\,-\frac{2(4+\ln x)}
{4x^2(4+\ln x)^{\frac32}}\\[4mm]
&\,\hspace{2pt}-\frac{1}
{4x^2(4+\ln x)^{\frac32}}\\[2mm]
=&\,-\frac{8+2\ln x+1}
{4x^2(4+\ln x)^{\frac32}}\\[2mm]
=&\,-\frac{9+2\ln x}
{4x^2(4+\ln x)^{\frac32}}.
\end{align*} d x 2 d 2 y = = = − 4 x 2 ( 4 + ln x ) 2 3 2 ( 4 + ln x ) − 4 x 2 ( 4 + ln x ) 2 3 1 − 4 x 2 ( 4 + ln x ) 2 3 8 + 2 ln x + 1 − 4 x 2 ( 4 + ln x ) 2 3 9 + 2 ln x .
解法二
思路
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也可以先平方：
y 2 = 4 + ln x . \begin{align*}
y^2=4+\ln x.
\end{align*} y 2 = 4 + ln x . 然后用 implicit differentiation。这个方法能少处理一些根号幂次。
答题过程
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Since
y = 4 + ln x , \begin{align*}
y=\sqrt{4+\ln x},
\end{align*} y = 4 + ln x , we have
y 2 = 4 + ln x . \begin{align*}
y^2=4+\ln x.
\end{align*} y 2 = 4 + ln x . Differentiate implicitly:
2 y d y d x = 1 x . \begin{align*}
2y\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,\frac1x.
\end{align*} 2 y d x d y = x 1 . So
d y d x = 1 2 x y . \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,\frac{1}{2xy}.
\end{align*} d x d y = 2 x y 1 . Differentiate
2 y d y d x = 1 x \begin{align*}
2y\frac{\mathrm{d}y}{\mathrm{d}x}
=\frac1x
\end{align*} 2 y d x d y = x 1 again:
2 y d 2 y d x 2 + 2 ( d y d x ) 2 = − 1 x 2 . \begin{align*}
2y\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
+2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2
=&\,-\frac1{x^2}.
\end{align*} 2 y d x 2 d 2 y + 2 ( d x d y ) 2 = − x 2 1 . Therefore
2 y d 2 y d x 2 = − 1 x 2 − 2 ( 1 2 x y ) 2 = − 1 x 2 − 1 2 x 2 y 2 . \begin{align*}
2y\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
=&\,-\frac1{x^2}
-2\left(\frac{1}{2xy}\right)^2\\[2mm]
=&\,-\frac1{x^2}
-\frac{1}{2x^2y^2}.
\end{align*} 2 y d x 2 d 2 y = = − x 2 1 − 2 ( 2 x y 1 ) 2 − x 2 1 − 2 x 2 y 2 1 . Divide by 2 y 2y 2 y
d 2 y d x 2 = − 1 2 x 2 y − 1 4 x 2 y 3 = − 2 y 2 + 1 4 x 2 y 3 . \begin{align*}
\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
=&\,-\frac{1}{2x^2y}
-\frac{1}{4x^2y^3}\\[2mm]
=&\,-\frac{2y^2+1}{4x^2y^3}.
\end{align*} d x 2 d 2 y = = − 2 x 2 y 1 − 4 x 2 y 3 1 − 4 x 2 y 3 2 y 2 + 1 . Since
y 2 = 4 + ln x , \begin{align*}
y^2=4+\ln x,
\end{align*} y 2 = 4 + ln x , we get
d 2 y d x 2 = − 2 ( 4 + ln x ) + 1 4 x 2 ( 4 + ln x ) 3 2 = − 9 + 2 ln x 4 x 2 ( 4 + ln x ) 3 2 . \begin{align*}
\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
=&\,-\frac{2(4+\ln x)+1}
{4x^2(4+\ln x)^{\frac32}}\\[2mm]
=&\,-\frac{9+2\ln x}
{4x^2(4+\ln x)^{\frac32}}.
\end{align*} d x 2 d 2 y = = − 4 x 2 ( 4 + ln x ) 2 3 2 ( 4 + ln x ) + 1 − 4 x 2 ( 4 + ln x ) 2 3 9 + 2 ln x .
(b)
解法一
思路
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Taylor expansion about x = 1 x=1 x = 1 y ( 1 ) y(1) y ( 1 ) y ′ ( 1 ) y'(1) y ′ ( 1 ) y ′ ′ ( 1 ) y''(1) y ′′ ( 1 )
答题过程
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At x = 1 x=1 x = 1
y ( 1 ) = 4 + ln 1 = 2. \begin{align*}
y(1)=\sqrt{4+\ln1}=2.
\end{align*} y ( 1 ) = 4 + ln 1 = 2. Also,
y ′ = 1 2 x ( 4 + ln x ) 1 2 . \begin{align*}
y'
=&\,\frac{1}{2x(4+\ln x)^{\frac12}}.
\end{align*} y ′ = 2 x ( 4 + ln x ) 2 1 1 . So
y ′ ( 1 ) = 1 2 ( 1 ) ( 4 ) 1 2 = 1 4 . \begin{align*}
y'(1)
=&\,\frac{1}{2(1)(4)^{\frac12}}\\[2mm]
=&\,\frac14.
\end{align*} y ′ ( 1 ) = = 2 ( 1 ) ( 4 ) 2 1 1 4 1 . From part (a),
y ′ ′ ( 1 ) = − 9 + 2 ln 1 4 ( 1 ) 2 ( 4 + ln 1 ) 3 2 = − 9 4 ⋅ 8 = − 9 32 . \begin{align*}
y''(1)
=&\,-\frac{9+2\ln1}
{4(1)^2(4+\ln1)^{\frac32}}\\[2mm]
=&\,-\frac{9}{4\cdot8}\\[2mm]
=&\,-\frac{9}{32}.
\end{align*} y ′′ ( 1 ) = = = − 4 ( 1 ) 2 ( 4 + ln 1 ) 2 3 9 + 2 ln 1 − 4 ⋅ 8 9 − 32 9 . Therefore
y = 2 + 1 4 ( x − 1 ) + 1 2 ! ( − 9 32 ) ( x − 1 ) 2 + ⋯ = 2 + 1 4 ( x − 1 ) − 9 64 ( x − 1 ) 2 + ⋯ . \begin{align*}
y
=&\,2+\frac14(x-1)\\[4mm]
&\,\hspace{2pt}+\frac{1}{2!}\Big(-\frac{9}{32}\Big)(x-1)^2+\cdots\\[4mm]
=&\,2+\frac14(x-1)-\frac{9}{64}(x-1)^2+\cdots.
\end{align*} y = = 2 + 4 1 ( x − 1 ) + 2 ! 1 ( − 32 9 ) ( x − 1 ) 2 + ⋯ 2 + 4 1 ( x − 1 ) − 64 9 ( x − 1 ) 2 + ⋯ .
解法二
思路
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令 h = x − 1 h=x-1 h = x − 1 ln x \ln x ln x ln ( 1 + h ) \ln(1+h) ln ( 1 + h ) h 2 h^2 h 2 4 4 4 ( 1 + u ) 1 2 (1+u)^{\frac12} ( 1 + u ) 2 1 u 2 u^2 u 2 u u u
答题过程
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Let h = x − 1 h=x-1 h = x − 1 x = 1 + h x=1+h x = 1 + h
ln ( 1 + h ) = h − h 2 2 + ⋯ , \begin{align*}
\ln(1+h)=h-\frac{h^2}{2}+\cdots,
\end{align*} ln ( 1 + h ) = h − 2 h 2 + ⋯ , write
y = 4 + ln ( 1 + h ) = 2 ( 1 + ln ( 1 + h ) 4 ) 1 2 . \begin{align*}
y=&\,\sqrt{4+\ln(1+h)}\\[4mm]
=&\,2\Big(1+\frac{\ln(1+h)}{4}\Big)^{\frac12}.
\end{align*} y = = 4 + ln ( 1 + h ) 2 ( 1 + 4 ln ( 1 + h ) ) 2 1 . Put
u = ln ( 1 + h ) 4 = h 4 − h 2 8 + ⋯ . \begin{align*}
u=&\,\frac{\ln(1+h)}{4}\\[4mm]
=&\,\frac{h}{4}-\frac{h^2}{8}+\cdots.
\end{align*} u = = 4 ln ( 1 + h ) 4 h − 8 h 2 + ⋯ . Up to and including the term in h 2 h^2 h 2
u 2 = ( h 4 ) 2 + ⋯ = h 2 16 + ⋯ . \begin{align*}
u^2=&\,\Big(\frac{h}{4}\Big)^2+\cdots\\[4mm]
=&\,\frac{h^2}{16}+\cdots.
\end{align*} u 2 = = ( 4 h ) 2 + ⋯ 16 h 2 + ⋯ . The binomial expansion gives
( 1 + u ) 1 2 = 1 + 1 2 u + 1 2 ( − 1 2 ) 2 ! u 2 + ⋯ = 1 + 1 2 u − 1 8 u 2 + ⋯ . \begin{align*}
(1+u)^{\frac12}
=&\,1+\frac12u\\[4mm]
&\,\hspace{2pt}+\frac{\frac12(-\frac12)}{2!}u^2+\cdots\\[4mm]
=&\,1+\frac12u-\frac18u^2+\cdots.
\end{align*} ( 1 + u ) 2 1 = = 1 + 2 1 u + 2 ! 2 1 ( − 2 1 ) u 2 + ⋯ 1 + 2 1 u − 8 1 u 2 + ⋯ . Therefore
y = 2 + u − 1 4 u 2 + ⋯ = 2 + h 4 − h 2 8 − 1 4 ( h 2 16 ) + ⋯ = 2 + h 4 − 8 h 2 64 − h 2 64 + ⋯ = 2 + h 4 − 9 h 2 64 + ⋯ . \begin{align*}
y=&\,2+u-\frac14u^2+\cdots\\[4mm]
=&\,2+\frac{h}{4}-\frac{h^2}{8}\\[4mm]
&\,\hspace{2pt}-\frac14\Big(\frac{h^2}{16}\Big)+\cdots\\[4mm]
=&\,2+\frac{h}{4}-\frac{8h^2}{64}-\frac{h^2}{64}+\cdots\\[4mm]
=&\,2+\frac{h}{4}-\frac{9h^2}{64}+\cdots.
\end{align*} y = = = = 2 + u − 4 1 u 2 + ⋯ 2 + 4 h − 8 h 2 − 4 1 ( 16 h 2 ) + ⋯ 2 + 4 h − 64 8 h 2 − 64 h 2 + ⋯ 2 + 4 h − 64 9 h 2 + ⋯ . Substituting h = x − 1 h=x-1 h = x − 1
y = 2 + 1 4 ( x − 1 ) − 9 64 ( x − 1 ) 2 + ⋯ . \begin{align*}
y=2+\frac14(x-1)-\frac{9}{64}(x-1)^2+\cdots.
\end{align*} y = 2 + 4 1 ( x − 1 ) − 64 9 ( x − 1 ) 2 + ⋯ . 