题目
Problem
Given that
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} = -\frac{9 + 2\ln x}{4x^2(4 + \ln x)^{\frac{3}{2}}}
\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)
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\left[
-x^{-2}(4+\ln x)^{-\frac12}\right.\\[2mm]
&\,\hspace{2pt}\left.x^{-1}\left(-\frac12\right)
(4+\ln x)^{-\frac32}\cdot\frac1x
\right]\\[2mm]
=&\,-\frac{1}{2x^2(4+\ln x)^{\frac12}}
-\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}}
-\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)
+\frac12\left(-\frac{9}{32}\right)(x-1)^2+\cdots\\[2mm]
=&\,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 + ⋯ . 