已知 y=ecos⁡−1xy = \mathrm{e}^{\cos^{-1} x}。

  1. (i)

    证明 (1−x2)d2ydx2−xdydx=y(1 - x^2)\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - x\frac{\mathrm{d}y}{\mathrm{d}x} = y。

  2. (ii)

    求 yy 的麦克劳林级数展开式，写出精确系数，直到并包括 x3x^3 项。

  3. (iii)

    由此求出 ecos⁡−1x1−x2\frac{\mathrm{e}^{\cos^{-1} x}}{\sqrt{1 - x^2}} 展开到并包括 x2x^2 的项，并估计曲线 y=ecos⁡−1xy = \mathrm{e}^{\cos^{-1} x} 在 x=0.5x = 0.5 处的切线斜率。

隐函数与参数方程求导前几项公式推导近似概念切线与法线English
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完整解法

6(i)

1. Differentiate $y$ with respect to $x$ using the chain rule.

y=ecos⁡−1xy = \mathrm{e}^{\cos^{-1} x}

dydx=ecos⁡−1x⋅(−11−x2)\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{\cos^{-1} x} \cdot \left( -\frac{1}{\sqrt{1 - x^2}} \right)

dydx=−y1−x2\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{y}{\sqrt{1 - x^2}}

2. Rearrange and square both sides to eliminate the square root, making the next differentiation simpler.

1−x2dydx=−y\sqrt{1 - x^2} \frac{\mathrm{d}y}{\mathrm{d}x} = -y

(1−x2)(dydx)2=y2(1 - x^2) \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = y^2

3. Differentiate implicitly with respect to $x$ and simplify to obtain the required expression.

Differentiating both sides with respect to xx using the product and chain rules:

(1−x2)(2dydxd2ydx2)+(−2x)(dydx)2=2ydydx(1 - x^2) \left( 2\frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} \right) + (-2x)\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 2y\frac{\mathrm{d}y}{\mathrm{d}x}

Dividing throughout by 2dydx2\frac{\mathrm{d}y}{\mathrm{d}x} (since dydx≠0\frac{\mathrm{d}y}{\mathrm{d}x} \neq 0):

(1−x2)d2ydx2−xdydx=y(1 - x^2)\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - x\frac{\mathrm{d}y}{\mathrm{d}x} = y (Shown)

答案

(Shown)

6(ii)

1. Evaluate $y$ and the first derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$ at $x=0$.

When x=0x = 0, y=ecos⁡−1(0)=eπ2y = \mathrm{e}^{\cos^{-1}(0)} = \mathrm{e}^{\frac{\pi}{2}}.

Differentiating y=ecos⁡−1xy = \mathrm{e}^{\cos^{-1} x} with respect to xx:
dydx=ecos⁡−1x⋅(−11−x2)\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{\cos^{-1} x} \cdot \left( \frac{-1}{\sqrt{1-x^2}} \right)

When x=0x = 0,
dydx=eπ2⋅(−11−0)=−eπ2\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{\frac{\pi}{2}} \cdot \left(\frac{-1}{\sqrt{1-0}}\right) = -\mathrm{e}^{\frac{\pi}{2}}

2. Use the differential equation from part (i) to evaluate the second derivative at $x=0$.

From part (i), we have:
(1−x2)d2ydx2−xdydx=y(1 - x^2)\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - x\frac{\mathrm{d}y}{\mathrm{d}x} = y

Substituting x=0x = 0, y=eπ2y = \mathrm{e}^{\frac{\pi}{2}}, and dydx=−eπ2\frac{\mathrm{d}y}{\mathrm{d}x} = -\mathrm{e}^{\frac{\pi}{2}}:
(1−0)d2ydx2−0=eπ2(1 - 0)\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - 0 = \mathrm{e}^{\frac{\pi}{2}}
d2ydx2=eπ2\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \mathrm{e}^{\frac{\pi}{2}}

3. Differentiate the identity from part (i) implicitly with respect to $x$ to evaluate the third derivative at $x=0$.

Differentiating (1−x2)d2ydx2−xdydx=y(1 - x^2)\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - x\frac{\mathrm{d}y}{\mathrm{d}x} = y with respect to xx applying the product rule:
(1−x2)d3ydx3+(−2x)d2ydx2−(xd2ydx2+(1)dydx)=dydx(1 - x^2)\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + (-2x)\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - \left(x\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + (1)\frac{\mathrm{d}y}{\mathrm{d}x}\right) = \frac{\mathrm{d}y}{\mathrm{d}x}
(1−x2)d3ydx3−3xd2ydx2−2dydx=0(1 - x^2)\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} - 3x\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - 2\frac{\mathrm{d}y}{\mathrm{d}x} = 0

Substituting x=0x = 0 and dydx=−eπ2\frac{\mathrm{d}y}{\mathrm{d}x} = -\mathrm{e}^{\frac{\pi}{2}}:
(1−0)d3ydx3−0−2(−eπ2)=0(1 - 0)\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} - 0 - 2\left(-\mathrm{e}^{\frac{\pi}{2}}\right) = 0
d3ydx3=−2eπ2\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} = -2\mathrm{e}^{\frac{\pi}{2}}

4. Construct the Maclaurin's series up to the term in $x^3$.

The Maclaurin's series is given by:
y=y(0)+y′(0)x+y′′(0)2!x2+y′′′(0)3!x3+…y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots
y=eπ2+(−eπ2)x+eπ22x2+−2eπ26x3+…y = \mathrm{e}^{\frac{\pi}{2}} + \left(-\mathrm{e}^{\frac{\pi}{2}}\right)x + \frac{\mathrm{e}^{\frac{\pi}{2}}}{2}x^2 + \frac{-2\mathrm{e}^{\frac{\pi}{2}}}{6}x^3 + \dots
y=eπ2−eπ2x+12eπ2x2−13eπ2x3+…y = \mathrm{e}^{\frac{\pi}{2}} - \mathrm{e}^{\frac{\pi}{2}}x + \frac{1}{2}\mathrm{e}^{\frac{\pi}{2}}x^2 - \frac{1}{3}\mathrm{e}^{\frac{\pi}{2}}x^3 + \dots

答案

y=eπ2−eπ2x+12eπ2x2−13eπ2x3y = \mathrm{e}^{\frac{\pi}{2}} - \mathrm{e}^{\frac{\pi}{2}}x + \frac{1}{2}\mathrm{e}^{\frac{\pi}{2}}x^2 - \frac{1}{3}\mathrm{e}^{\frac{\pi}{2}}x^3

6(iii)

1. Derive the Maclaurin series for $y$ up to $x^3$ as required from part (ii).

From part (ii):
At x=0x=0, y(0)=ecos⁡−10=eπ2y(0) = \mathrm{e}^{\cos^{-1} 0} = \mathrm{e}^{\frac{\pi}{2}}.
dydx=−ecos⁡−1x1−x2  ⟹  y′(0)=−eπ2\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{\mathrm{e}^{\cos^{-1} x}}{\sqrt{1 - x^2}} \implies y'(0) = -\mathrm{e}^{\frac{\pi}{2}}.
Using the given equation (1−x2)y′′−xy′=y(1 - x^2)y'' - xy' = y, substituting x=0x=0 yields y′′(0)=eπ2y''(0) = \mathrm{e}^{\frac{\pi}{2}}.
Differentiating the given equation implicitly yields (1−x2)y′′′−3xy′′−2y′=0(1-x^2)y''' - 3xy'' - 2y' = 0, giving y′′′(0)=−2eπ2y'''(0) = -2\mathrm{e}^{\frac{\pi}{2}}.
Thus, the Maclaurin series is y≈eπ2(1−x+12x2−13x3)y \approx \mathrm{e}^{\frac{\pi}{2}} \left( 1 - x + \frac{1}{2}x^2 - \frac{1}{3}x^3 \right).

2. Relate the target expression to $\frac{\mathrm{d}y}{\mathrm{d}x}$ and find its expansion by differentiating the series.

Observe that the required expression ecos⁡−1x1−x2\frac{\mathrm{e}^{\cos^{-1} x}}{\sqrt{1 - x^2}} is exactly −dydx-\frac{\mathrm{d}y}{\mathrm{d}x}.
Differentiating the Maclaurin series for yy term-by-term:
dydx≈eπ2(−1+x−x2)\frac{\mathrm{d}y}{\mathrm{d}x} \approx \mathrm{e}^{\frac{\pi}{2}} \left( -1 + x - x^2 \right)
Multiplying by −1-1 gives the required expansion up to x2x^2:
ecos⁡−1x1−x2≈eπ2(1−x+x2)\frac{\mathrm{e}^{\cos^{-1} x}}{\sqrt{1 - x^2}} \approx \mathrm{e}^{\frac{\pi}{2}} \left( 1 - x + x^2 \right)

3. Estimate the gradient of the tangent at $x = 0.5$.

The gradient of the tangent to the curve is given by dydx\frac{\mathrm{d}y}{\mathrm{d}x}.
Using the series for dydx\frac{\mathrm{d}y}{\mathrm{d}x} derived above:
Gradient≈eπ2(−1+x−x2)\text{Gradient} \approx \mathrm{e}^{\frac{\pi}{2}} \left( -1 + x - x^2 \right)
Substitute x=0.5x = 0.5:
Gradient≈eπ2(−1+0.5−(0.5)2)\text{Gradient} \approx \mathrm{e}^{\frac{\pi}{2}} \left( -1 + 0.5 - (0.5)^2 \right)
Gradient≈eπ2(−0.75)=−34eπ2\text{Gradient} \approx \mathrm{e}^{\frac{\pi}{2}} (-0.75) = -\frac{3}{4}\mathrm{e}^{\frac{\pi}{2}}

答案

Expansion: eπ2(1−x+x2)\mathrm{e}^{\frac{\pi}{2}}(1 - x + x^2)

Gradient estimate: −34eπ2-\frac{3}{4}\mathrm{e}^{\frac{\pi}{2}}

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