第 12 类 RD Sharma 解决方案 - 第 11 章微分 - 练习 11.2 |设置 3

问题 50. 微分 y = $3 e^{- 3x} \log \left( 1 + x \right)$关于 x。

解决方案：

We have,

y = $3 e^{- 3x} \log \left( 1 + x \right)$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left(3 e^{- 3x} \log \left( 1 + x \right) \right)$

$\frac{d y}{d x} = 3\frac{d}{dx}\left[ e^{- 3x} \log\left( 1 + x \right) \right]$

On using product rule and chain rule, we get

$\frac{d y}{d x} = 3\left\{ e^{- 3x} \frac{1}{1 + x} + \log\left( 1 + x \right)\left( - 3 e^{- 3x} \right) \right\}$

$\frac{d y}{d x} = 3\left\{ \frac{e^{- 3x}}{1 + x} - 3 e^{- 3x} \log\left( 1 + x \right) \right\}$

$\frac{d y}{d x} = 3 e^{- 3x} \left\{ \frac{1}{1 + x} - 3 \log\left( 1 + x \right) \right\}$

问题 51. 微分 y = $\frac{x^2 + 2}{\sqrt{\cos x}}$关于 x。

解决方案：

We have,

y = $\frac{x^2 + 2}{\sqrt{\cos x}}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left(\frac{x^2 + 2}{\sqrt{\cos x}}\right)$

On using quotient rule and chain rule, we get

$\frac{d y}{d x} = \frac{\sqrt{\cos x}\frac{d}{dx}\left( x^2 + 2 \right) - \left( x^2 + 2 \right)\frac{d}{dx}\left( \sqrt{\cos x} \right)}{\left( \sqrt{\cos x} \right)^2}$

$\frac{d y}{d x} = \frac{2x\sqrt{\cos x} - \left( x^2 + 2 \right)\left( - \frac{1}{2}\frac{\sin x}{\sqrt{\cos x}} \right)}{\cos x}$

$\frac{d y}{d x} = \frac{2x\sqrt{\cos x} + \frac{\left( x^2 + 2 \right)\sin x}{2\sqrt{\cos x}}}{\cos x}$

$\frac{d y}{d x} = \frac{4x \cos x + \left( x^2 + 2 \right)\sin x}{2 \left( \cos x \right)^\frac{3}{2}}$

$\frac{d y}{d x} = \frac{2x}{\sqrt{\cos x}} + \frac{1}{2}\frac{\left( x^2 + 2 \right)\sin x}{\left( \cos x \right)^\frac{3}{2}}$

$\frac{d y}{d x} = \frac{1}{\sqrt{\cos x}}\left\{ 2x + \frac{1}{2}\frac{\left( x^2 + 2 \right)\sin x}{\cos x} \right\}$

$\frac{d y}{d x} = \frac{1}{\sqrt{\cos x}}\left\{ 2x + \frac{\left( x^2 + 2 \right)\tan x}{2} \right\}$

问题 52. 微分 y = $\frac{x^2 \left( 1 - x^2 \right)}{\cos 2x}$关于 x。

解决方案：

We have,

y = $\frac{x^2 \left( 1 - x^2 \right)}{\cos 2x}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left(\frac{x^2 \left( 1 - x^2 \right)}{\cos 2x}\right)$

On using quotient rule and chain rule, we get

$\frac{d y}{d x} = \frac{\cos2x\frac{d}{dx}\left\{ x^2 \left( 1 - x^2 \right)^3 \right\} - x^2 \left( 1 - x^2 \right)^3 \frac{d}{dx}\cos2x}{\cos^2 2x}$

$\frac{d y}{d x} = \frac{\cos2x\left\{ x^2 \frac{d}{dx} \left( 1 - x^2 \right)^3 + \left( 1 - x^2 \right)^3 \frac{d}{dx}\left( x^2 \right) \right\} - x^2 \left( 1 - x^2 \right)^3 \left( - 2\sin2x \right)}{\cos^2 2x}$

$\frac{d y}{d x} = \frac{\cos2x\left\{ - 6 x^3 \left( 1 - x^2 \right)^2 + \left( 1 - x^2 \right)^3 2x \right\} + 2 x^2 \left( 1 - x^2 \right)^3 \sin2x}{\cos^2 2x}$

$\frac{d y}{d x} = \frac{2x \left( 1 - x^2 \right)^2}{\cos2x} - \frac{6 x^3 \left( 1 - x^2 \right)^2}{\cos2x} + \frac{2 x^2 \left( 1 - x^2 \right)^3 \sin2x}{\cos^2 2x}$

$\frac{d y}{d x} = 2x\left( 1 - x^2 \right)\sec2x\left\{ 1 - 4 x^2 + x\left( 1 - x^2 \right)\tan2x \right\}$

问题 53. 微分 y = $\log\left\{ \cot\left( \frac{\pi}{4} + \frac{x}{2} \right) \right\}$关于 x。

解决方案：

We have,

y = $\log\left\{ \cot\left( \frac{\pi}{4} + \frac{x}{2} \right) \right\}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left(\log\left\{ \cot\left( \frac{\pi}{4} + \frac{x}{2} \right) \right\}\right)$

On using chain rule, we get

$\frac{d y}{d x} = \frac{- {cosec}^2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}{2\cot\left( \frac{\pi}{4} + \frac{x}{2} \right)}$

$\frac{d y}{d x} = \frac{- 1}{2\cos\left( \frac{\pi}{4} + \frac{x}{2} \right)\sin\left( \frac{\pi}{4} + \frac{x}{2} \right)}$

$\frac{d y}{d x} = \frac{- 1}{\sin\left( \frac{\pi}{2} + x \right)}$

$\frac{d y}{d x} = \frac{- 1}{\cos x}$

$\frac{d y}{d x} = - \sec x$

问题 54. 微分 y = $e^{ax} \sec x \tan2x$关于 x。

解决方案：

We have,

y = $e^{ax} \sec x \tan2x$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( e^{ax} sec x \tan2x \right)$

On using chain rule, we get

$\frac{d y}{d x} = e^{ax} \frac{d}{dx}\left\{ \sec x \tan2x \right\} + \sec x \tan2x\frac{d}{dx}\left\{ e^{ax} \right\}$

$\frac{d y}{d x} = e^{ax} \left[ \sec x \tan x \tan2x + 2 \sec^2 2x\sec x \right] + a e^{ax} \sec x \tan2x$

$\frac{d y}{d x} = e^{ax} \left[ \sec x \tan x \tan2x + 2 \sec^2 2x\sec x \right] + a e^{ax} sec x \tan2x$

$\frac{d y}{d x} = a e^{ax} \sec x \tan2x + e^{ax} \sec x \tan x \tan2x + 2 e^{ax} \sec x \sec^2 2x$

$\frac{d y}{d x} = e^{ax} \sec x\left\{ a \tan2x + \tan x \tan2x + 2 \sec^2 2x \right\}$

问题 55. 微分 y = $\log \left( \cos x^2 \right)$关于 x。

解决方案：

We have,

y = $\log \left( \cos x^2 \right)$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( \log \left( \cos x^2 \right) \right)$

On using chain rule, we get

$\frac{d y}{d x} = \frac{d}{dx}\left\{ \log\left( \cos x^2 \right) \right\}$

$\frac{d y}{d x} = \frac{- 2x \sin x^2}{\cos x^2}$

$\frac{d y}{d x} = - 2x \tan x^2$

问题 56. 微分 y = $\cos \left( \log x \right)^2$关于 x。

解决方案：

We have,

y = $\cos \left( \log x \right)^2$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( \log \left( \cos x^2 \right) \right)$

On using chain rule, we get

$\frac{d y}{d x} = - \sin \left( \log x \right)^2 \frac{d}{dx} \left( \log x \right)^2$

$\frac{d y}{d x} = - \sin \left( \log x \right)^2 \frac{2\log x}{x}$

$\frac{d y}{d x} = \frac{- 2\log x \sin \left( \log x \right)^2}{x}$

问题 57. 微分 y = $\log \sqrt{\frac{x - 1}{x + 1}}$关于 x。

解决方案：

We have,

y = $\log \sqrt{\frac{x - 1}{x + 1}}$

y = $\log \left( \frac{x - 1}{x + 1} \right)^\frac{1}{2}$

y = $\frac{1}{2}\log \left( \frac{x - 1}{x + 1} \right)$

y = $\frac{1}{2}\left[ \log\left( x - 1 \right) - \log\left( x + 1 \right) \right]$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( \frac{1}{2}\left[ \log\left( x - 1 \right) - \log\left( x + 1 \right) \right] \right)$

$\frac{d y}{d x} = \frac{1}{2}\left[ \frac{d}{dx}\left\{ \log\left( x - 1 \right) \right\} - \frac{d}{dx}\left\{ \log\left( x + 1 \right) \right\} \right]$

$\frac{d y}{d x} = \frac{1}{2}\left( \frac{1}{x - 1} - \frac{1}{x + 1} \right)$

$\frac{d y}{d x} = \frac{1}{2}\left( \frac{2}{x^2 - 1} \right)$

$\frac{d y}{d x} = \frac{1}{x^2 - 1}$

问题 58. 如果$y = \log \left\{ \sqrt{x - 1} - \sqrt{x + 1} \right\}$ ， 显示$\frac{dy}{dx} = \frac{- 1}{2\sqrt{x^2 - 1}}$ .

解决方案：

We have,

y = $\log \left\{ \sqrt{x - 1} - \sqrt{x + 1} \right\}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\log\left( \sqrt{x - 1} - \sqrt{x + 1} \right)$

On using chain rule, we get

$\frac{d y}{d x} = \frac{1}{\left( \sqrt{x - 1} - \sqrt{x + 1} \right)}\frac{d}{dx}\left( \sqrt{x - 1} - \sqrt{x + 1} \right)$

$\frac{d y}{d x} = \frac{1}{\left( \sqrt{x - 1} - \sqrt{x + 1} \right)}\left[ \frac{d}{dx}\sqrt{x - 1} - \frac{d}{dx}\sqrt{x + 1} \right]$

$\frac{d y}{d x} = \frac{1}{\left( \sqrt{x - 1} - \sqrt{x + 1} \right)}\left[ \frac{1}{2} \left( x - 1 \right)^\frac{- 1}{2} - \frac{1}{2} \left( x + 1 \right)^\frac{- 1}{2} \right]$

$\frac{d y}{d x} = \frac{1}{2}\frac{1}{\left( \sqrt{x - 1} - \sqrt{x + 1} \right)}\left( \frac{1}{\sqrt{x - 1}} - \frac{1}{\sqrt{x + 1}} \right)$

$\frac{d y}{d x} = \frac{1}{2}\frac{1}{\left( \sqrt{x - 1} - \sqrt{x + 1} \right)}\left\{ \frac{- \left( \sqrt{x - 1} - \sqrt{x + 1} \right)}{\left( \sqrt{x - 1} \right)\left( \sqrt{x + 1} \right)} \right\}$

$\frac{d y}{d x} = \frac{- 1}{2}\left( \frac{1}{\left( \sqrt{x - 1} \right)\left( \sqrt{x + 1} \right)} \right)$

$\frac{d y}{d x} = \frac{- 1}{2\sqrt{x^2 - 1}}$

Hence, proved.

问题 59. 如果 y = $\sqrt{x + 1} + \sqrt{x - 1}$ ， 证明$\sqrt{x^2 - 1}\frac{dy}{dx} = \frac{1}{2}y$ .

解决方案：

We have,

y = $\sqrt{x + 1} + \sqrt{x - 1}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\log\left( \sqrt{x - 1} - \sqrt{x + 1} \right)$

$\frac{d y}{d x} = \frac{d}{dx}\left( \sqrt{x + 1} \right) + \frac{d}{dx}\left( \sqrt{x - 1} \right)$

$\frac{d y}{d x} = \frac{1}{2} \left( x + 1 \right)^\frac{- 1}{2} + \frac{1}{2} \left( x - 1 \right)^\frac{- 1}{2}$

$\frac{d y}{d x} = \frac{1}{2}\left( \frac{1}{\sqrt{x + 1}} + \frac{1}{\sqrt{x - 1}} \right)$

$\frac{d y}{d x} = \frac{1}{2}\left( \frac{\sqrt{x - 1} + \sqrt{x + 1}}{\left( \sqrt{x + 1} \right)\left( \sqrt{x - 1} \right)} \right)$

$\frac{d y}{d x} = \frac{1}{2}\left( \frac{y}{\left( \sqrt{x^2 - 1} \right)} \right)$

$\left( \sqrt{x^2 - 1} \right)\frac{d y}{d x} = \frac{1}{2}y$

Hence proved.

问题 60. 如果 y = $\frac{x}{x + 2}$ ， 证明$x\frac{dy}{dx} = \left( 1 - y \right) y$ .

解决方案：

We have,

y = $\frac{x}{x + 2}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( \frac{x}{x + 2} \right)$

$\frac{d y}{d x} = \frac{\left( x + 2 \right)\frac{d}{dx}\left( x \right) - x\frac{d}{dx}\left( x + 2 \right)}{\left( x + 2 \right)^2}$

$\frac{d y}{d x} = \frac{x + 2 - x}{\left( x + 2 \right)^2}$

$\frac{d y}{d x} = \frac{x + 2}{\left( x + 2 \right)^2} - \frac{x}{\left( x + 2 \right)^2}$

As $x + 2 = \frac{x}{y}$, we have

$\frac{d y}{d x} = \frac{1}{x + 2} - \frac{x y^2}{x^2}$

$\frac{d y}{d x} = \frac{y}{x} - \frac{y^2}{x}$

$\frac{d y}{d x} = \frac{1}{x}y\left( 1 - y \right)$

$x\frac{d y}{d x} = \left( 1 - y \right)y$

Hence proved.

问题 61. 如果 y = $\log \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)$ ， 证明$\frac{dy}{dx} = \frac{x - 1}{2x \left( x + 1 \right)}$ .

解决方案：

We have,

y = $\log \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\log\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)$

$\frac{d y}{d x} = \frac{1}{\sqrt{x} + \frac{1}{\sqrt{x}}}\frac{d}{dx}\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)$

$\frac{d y}{d x} = \frac{\sqrt{x}}{x + 1}\left( \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} \right)$

$\frac{d y}{d x} = \frac{1}{2}\frac{\sqrt{x}}{x + 1}\left( \frac{x - 1}{x\sqrt{x}} \right)$

$\frac{d y}{d x} = \frac{x - 1}{2x\left( x + 1 \right)}$

Hence proved.

问题 62. 如果 y = $\log \sqrt{\frac{1 + \tan x}{1 - \tan x}}$ ， 证明$\frac{dy}{dx} = \sec 2x$ .

解决方案：

We have,

y = $\log \sqrt{\frac{1 + \tan x}{1 - \tan x}}$

y = $\log \left\{ \frac{1 + \tan x}{1 - \tan x} \right\}^\frac{1}{2}$

y = $\frac{1}{2}\log\left\{ \frac{1 + \tan x}{1 - \tan x} \right\}$

y = $\frac{1}{2}\left\{ \log\left( 1 + \tan x \right) - \log\left( 1 - \tan x \right) \right\}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\log\left(\frac{1}{2}\left\{ \log\left( 1 + \tan x \right) - \log\left( 1 - \tan x \right) \right\} \right)$

$\frac{d y}{d x} = \frac{1}{2}\left\{ \frac{d}{dx}\left\{ \log\left( 1 + \tan x \right) \right\} - \frac{d}{dx}\left\{ \log\left( 1 - \tan x \right) \right\} \right\}$

$\frac{d y}{d x} = \frac{1}{2}\left\{ \frac{1}{1 + \tan x } \times \frac{d}{dx}\left( 1 + \tan x \right) - \frac{1}{1 - \tan x} \times \frac{d}{dx}\left( 1 - \tan x \right) \right\}$

$\frac{d y}{d x} = \frac{1}{2}\left\{ \frac{1}{1 + \tan x}\left( 0 + \sec^2 x \right) - \frac{1}{1 - \tan x}\left( 0 - \sec^2 x \right) \right\}$

$\frac{d y}{d x}= \frac{1}{2}\left\{ \frac{\sec^2 x}{1 + \tan x} + \frac{\sec^2 x}{1 - \tan x} \right\}$

$\frac{d y}{d x} = \frac{1}{2} \sec^2 x\left\{ \frac{1 - \tan x + 1 + \tan x}{1 - \tan^2 x} \right\}$

$\frac{d y}{d x} = \frac{1}{2} \sec^2 x\left( \frac{2}{1 - \tan^2 x} \right)$

$\frac{d y}{d x} = \frac{\sec^2 x}{1 - \tan^2 x}$

$\frac{d y}{d x} = \frac{1 + \tan^2 x}{1 - \tan^2 x}$

$\frac{d y}{d x} = \frac{1}{\frac{1 - \tan^2 x}{1 + \tan^2 x}}$

$\frac{d y}{d x} = \frac{1}{\cos2x}$

$\frac{d y}{d x} = \sec2x$

Hence proved.

问题 63. 如果 y = $\sqrt{x} + \frac{1}{\sqrt{x}}$ ， 证明$2 x\frac{dy}{dx} = \sqrt{x} - \frac{1}{\sqrt{x}}$ .

解决方案：

We have,

y = $\sqrt{x} + \frac{1}{\sqrt{x}}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)$

$\frac{d y}{d x} = \frac{d}{dx}\left( \sqrt{x} \right) + \frac{d}{dx}\left( \frac{1}{\sqrt{x}} \right)$

$\frac{d y}{d x} = \frac{1}{2\sqrt{x}} + \left( \frac{- \frac{1}{2\sqrt{x}}}{x} \right)$

$\frac{d y}{d x} = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}}$

$\frac{d y}{d x} = \frac{x - 1}{2x\sqrt{x}}$

$2x\frac{d y}{d x} = \frac{x - 1}{\sqrt{x}}$

$2x\frac{d y}{d x} = \frac{x}{\sqrt{x}} - \frac{1}{\sqrt{x}}$

$2x\frac{d y}{d x} = \sqrt{x} - \frac{1}{\sqrt{x}}$

Hence proved.

问题 64. 如果 y = $\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}$ ， 证明$\left( 1 - x^2 \right) \frac{dy}{dx} = x + \frac{y}{x}$ .

解决方案：

We have,

y = $\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}} \right)$

$\frac{d y}{d x} = \left[ \frac{\sqrt{1 - x^2}\frac{d}{dx}\left( x \sin^{- 1} x \right) - \left( x \sin^{- 1} x \right)\frac{d}{dx}\left( \sqrt{1 - x^2} \right)}{\left( \sqrt{1 - x^2} \right)^2} \right]$

$\frac{d y}{d x} = \left[ \frac{\sqrt{1 - x^2}\left\{ x\frac{d}{dx}\left( \sin^{- 1} x \right) + \sin^{- 1} x\frac{d}{dx}\left( x \right) \right\} - \left( x \sin^{- 1} x \right)\frac{1}{2\sqrt{1 - x^2}}\frac{d}{dx}\left( 1 - x^2 \right)}{\left( 1 - x^2 \right)} \right]$

$\frac{d y}{d x} = \left[ \frac{\sqrt{1 - x^2}\left\{ \frac{x}{\sqrt{1 - x^2}} + \sin^{- 1} x \right\} - \frac{x \sin^{- 1} x\left( - 2x \right)}{2\sqrt{1 - x^2}}}{\left( 1 - x^2 \right)} \right]$

$\frac{d y}{d x} = \left[ \frac{x + \sqrt{1 - x^2} \sin^{- 1} x + \frac{x^2 \sin^{- 1} x}{\sqrt{1 - x^2}}}{\left( 1 - x^2 \right)} \right]$

$\left( 1 - x^2 \right)\frac{d y}{d x} = x + \frac{\sqrt{1 - x^2} \sin^{- 1} x}{1} + \frac{x^2 \sin^{- 1} x}{\sqrt{1 - x^2}}$

$\left( 1 - x^2 \right)\frac{d y}{d x} = x + \left( \frac{\left( 1 - x^2 \right) \sin^{- 1} x + x^2 \sin^{- 1} x}{\sqrt{1 - x^2}} \right)$

$\left( 1 - x^2 \right)\frac{d y}{d x} = x + \left( \frac{\sin^{- 1} x - x^2 \sin^{- 1} x + x^2 \sin^{- 1} x}{\sqrt{1 - x^2}} \right)$

$\left( 1 - x^2 \right)\frac{d y}{d x} = x + \left( \frac{\sin^{- 1} x}{\sqrt{1 - x^2}} \right)$

As we have $y = \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}$, we get

$\left( 1 - x^2 \right)\frac{d y}{d x} = x + \frac{y}{x}$

Hence proved.

问题 65. 如果 y = $\frac{e^x - e^{- x}}{e^x + e^{- x}}$ ， 证明$\frac{dy}{dx} = 1 - y^2$ .

解决方案：

We have,

y = $\frac{e^x - e^{- x}}{e^x + e^{- x}}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( \frac{e^x - e^{- x}}{e^x + e^{- x}} \right)$

$\frac{d y}{d x} = \left[ \frac{\left( e^x + e^{- x} \right)\frac{d}{dx}\left( e^x - e^{- x} \right) - \left( e^x - e^{- x} \right)\frac{d}{dx}\left( e^x + e^{- x} \right)}{\left( e^x + e^{- x} \right)^2} \right]$

$\frac{d y}{d x} = \left[ \frac{\left( e^x + e^{- x} \right)\left\{ e^x - e^{- x} \frac{d}{dx}\left( - x \right) \right\} - \left( e^x - e^{- x} \right)\left\{ e^x + e^{- x} \frac{d}{dx}\left( - x \right) \right\}}{\left( e^x + e^{- x} \right)^2} \right]$

$\frac{d y}{d x} = \left[ \frac{\left( e^x + e^{- x} \right)\left( e^x + e^{- x} \right) - \left( e^x - e^{- x} \right)\left( e^x - e^{- x} \right)}{\left( e^x + e^{- x} \right)^2} \right]$

$\frac{d y}{d x} = \frac{\left( e^x + e^{- x} \right)^2 - \left( e^x - e^{- x} \right)^2}{\left( e^x + e^{- x} \right)^2}$

$\frac{d y}{d x} = 1 - \frac{\left( e^x - e^{- x} \right)^2}{\left( e^x + e^{- x} \right)^2}$

$\frac{d y}{d x} = 1 - \left( \frac{e^x - e^{- x}}{e^x + e^{- x}} \right)^2$

As y = $\frac{e^x - e^{- x}}{e^x + e^{- x}}$, we have

$\frac{d y}{d x} = 1 - y^2$

Hence proved.

问题 66. 如果 y = $\left( x - 1 \right) \log \left( x - 1 \right) - \left( x + 1 \right) \log \left( x + 1 \right)$ ， 证明$\frac{dy}{dc} = \log \left( \frac{x - 1}{1 + x} \right)$ .

解决方案：

We have,

y = $\left( x - 1 \right) \log \left( x - 1 \right) - \left( x + 1 \right) \log \left( x + 1 \right)$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ \left( x - 1 \right) \log\left( x - 1 \right) - \left( x + 1 \right) \log\left( x + 1 \right) \right]$

$\frac{d y}{d x} = \left[ \left( x - 1 \right)\frac{d}{dx}\log\left( x - 1 \right) + \log\left( x - 1 \right)\frac{d}{dx}\left( x - 1 \right) \right] - \left[ \left( x + 1 \right)\frac{d}{dx}\log\left( x + 1 \right) + \log\left( x + 1 \right)\frac{d}{dx}\left( x + 1 \right) \right]$

$\frac{d y}{d x} = \left[ \left( x - 1 \right) \times \frac{1}{\left( x - 1 \right)}\frac{d}{dx}\left( x - 1 \right) + \log\left( x - 1 \right) \times \left( 1 \right) \right] - \left[ \left( x + 1 \right) \times \frac{1}{\left( x + 1 \right)} \times \frac{d}{dx}\left( x + 1 \right) + \log\left( x + 1 \right)\left( 1 \right) \right]$

$\frac{d y}{d x} = \left[ 1 + \log\left( x - 1 \right) \right] - \left[ 1 + \log\left( x + 1 \right) \right]$

$\frac{d y}{d x} = \log\left( x - 1 \right) - \log\left( x + 1 \right)$

$\frac{d y}{d x} = \log\frac{\left( x - 1 \right)}{\left( x + 1 \right)}$

Hence proved.

问题 67. 如果 y = $e^x \cos x$ ， 证明$\frac{dy}{dx} = \sqrt{2} e^x \cdot \cos \left( x + \frac{\pi}{4} \right)$ .

解决方案：

We have,

y = $e^x \cos x$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( e^x \cos x \right)$

$\frac{d y}{d x} = e^x \frac{d}{dx}\cos x + \cos x\frac{d}{dx} e^x$

$\frac{d y}{d x} = e^x \left( - \sin x \right) + e^x \cos x$

$\frac{d y}{d x} = e^x \left( \cos x - \sin x \right)$

$\frac{d y}{d x} = \sqrt{2} e^x \left( \frac{\cos x}{\sqrt{2}} - \frac{\sin x}{\sqrt{2}} \right)$

$\frac{d y}{d x} = \sqrt{2} e^x \left( \cos\frac{\pi}{4}\cos x - \sin\frac{\pi}{4}\sin x \right)$

$\frac{d y}{d x} = \sqrt{2} e^x \cos\left( x + \frac{\pi}{4} \right)$

Hence proved.

问题 68. 如果 y = $\frac{1}{2} \log \left( \frac{1 - \cos 2x }{1 + \cos 2x} \right)$ ， 证明$\frac{ dy }{ dx } = 2 \text{cosec }2x$ .

解决方案：

We have,

y = $\frac{1}{2} \log \left( \frac{1 - \cos 2x }{1 + \cos 2x} \right)$

y = $\frac{1}{2}\log\left( \frac{2 \sin^2 x}{2 \cos^2 x} \right)$

y = $\frac{1}{2}\log\left( \tan^2 x \right)$

y = $\frac{2}{2}\log \tan x$

y = $\log \tan x$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left(\log \tan x\right)$

$\frac{d y}{d x} = \frac{1}{\tan x} \times \frac{d}{dx}\left( \tan x \right)$

$\frac{d y}{d x} = \frac{\sec^2 x}{\tan x}$

$\frac{d y}{d x} = \frac{1}{\cos^2 x \times \frac{\sin x}{\cos x}}$

$\frac{d y}{d x} = \frac{1}{\sin x \cos x}$

$\frac{d y}{d x} = \frac{2}{2\sin x \cos x}$

$\frac{d y}{d x} = \frac{2}{\sin2x}$

$\frac{d y}{d x} = 2\cosec x$

Hence proved.

问题 69. 如果 y = $x \sin^{- 1} x + \sqrt{1 - x^2}$ ， 证明$\frac{dy}{dx} = \sin^{- 1} x$ .

解决方案：

We have,

y = $x \sin^{- 1} x + \sqrt{1 - x^2}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ x \sin^{- 1} x + \sqrt{1 - x^2} \right]$

$\frac{d y}{d x} = \frac{d}{dx}\left( x \sin^{- 1} x \right) + \frac{d}{dx}\left( \sqrt{1 - x^2} \right)$

$\frac{d y}{d x} = \left[ x \frac{d}{dx} \sin^{- 1} x + \sin^{- 1} x\frac{d}{dx}\left( x \right) \right] + \frac{1}{2\sqrt{1 - x^2}}\frac{d}{dx}\left( 1 - x^2 \right)$

$\frac{d y}{d x} = \left[ \frac{x}{\sqrt{1 - x^2}} + \sin^{- 1} x \right] - \frac{2x}{2\sqrt{1 - x^2}}$

$\frac{d y}{d x} = \frac{x}{\sqrt{1 - x^2}} + \sin^{- 1} x - \frac{x}{\sqrt{1 - x^2}}$

$\frac{d y}{d x} = \sin^{- 1} x$

Hence proved.

问题 70. 如果 y = $\sqrt{x^2 + a^2}$ ， 证明$y\frac{dy}{dx} - x = 0$ .

解决方案：

We have,

y = $\sqrt{x^2 + a^2}$

On squaring both sides we get,

$y^2 = x^2 + a^2$

On differentiating both sides with respect to x we get,

$2y\frac{d y}{d x} = \frac{d}{dx}\left( x^2 + a^2 \right)$

$2y\frac{d y}{d x} = \left( 2x \right)$

$2y\frac{d y}{d x} = 2x$

$y\frac{d y}{d x} = x$

$y\frac{d y}{d x} - x = 0$

Hence proved.

问题 71. 如果 y = $e^x + e^{- x}$ ， 证明$\frac{dy}{dx} = \sqrt{y^2 - 4}$ .

解决方案：

We have,

y = $e^x + e^{- x}$

On differentiating both sides with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( e^x + e^{- x} \right)$

$\frac{d y}{d x} = \frac{d}{dx} e^x + \frac{d}{dx} e^{- x}$

On using chain rule, we get 

$\frac{d y}{d x} = e^x + e^{- x} \frac{d}{dx}\left( - x \right)$

$\frac{d y}{d x} = e^x + e^{- x} \left( - 1 \right)$

$\frac{d y}{d x} = \left( e^x - e^{- x} \right)$

$\frac{d y}{d x} = \sqrt{\left( e^x + e^{- x} \right)^2 - 4 e^x \times e^{- x}}$

As $e^x + e^{- x} = y$, we get

$\frac{d y}{d x} = \sqrt{y^2 - 4}$

Hence proved.

问题 72. 如果 y = $\sqrt{a^2 - x^2}$ ， 证明$y\frac{dy}{dx} + x = 0$ .

解决方案：

We have,

y = $\sqrt{a^2 - x^2}$

On squaring both sides we get,

$y^2 = a^2 - x^2$

On differentiating both sides with respect to x we get,

$2y\frac{d y}{d x} = \frac{d}{dx}\left( a^2 - x^2 \right)$

$2y\frac{d y}{d x} = 0 - 2x$

$y\frac{d y}{d x} = - x$

$y\frac{d y}{d x} + x = 0$

Hence proved.

问题 73.如果 xy = 4，证明$x\left( \frac{dy}{dx} + y^2 \right) = 3 y$ .

解决方案：

We have,

xy = 4

y = $\frac{4}{x}$

On differentiating both sides with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( \frac{4}{x} \right)$

$\frac{d y}{d x} = 4\frac{d}{dx}\left( x^{- 1} \right)$

$\frac{d y}{d x} = 4\left( - 1 \times x^{- 1 - 1} \right)$

$\frac{d y}{d x} = 4\left( - \frac{1}{x^2} \right)$

$\frac{d y}{d x} = \frac{- 4}{x^2}$

As x = $\frac{4}{y}$, we get

$\frac{d y}{d x} = - \frac{4}{\left( \frac{4}{y} \right)^2}$

$\frac{d y}{d x} = - \frac{4 y^2}{16}$

$\frac{d y}{d x} = - \frac{y^2}{4}$

$4\frac{d y}{d x} = - y^2$

$4\frac{d y}{d x} = 3 y^2 - 4 y^2$

$4\frac{d y}{d x} + 4 y^2 = 3 y^2$

$4\left( \frac{d y}{d x} + y^2 \right) = 3 y^2$

On dividing both sides by x, we get

$\frac{4}{x}\left( \frac{d y}{d x} + y^2 \right) = \frac{3 y^2}{x}$

$y\left( \frac{d y}{d x} + y^2 \right) = \frac{3 y^2}{x}$

$x\left( \frac{d y}{d x} + y^2 \right) = \frac{3 y^2}{y}$

$x\left( \frac{d y}{d x} + y^2 \right) = 3y$

Hence proved.

问题 74. 证明$\frac{d}{dx} \left\{ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{- 1} \frac{x}{a} \right\} = \sqrt{a^2 - x^2}$ .

解决方案：

We have,

L.H.S. = $\frac{d}{dx} \left\{ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{- 1} \frac{x}{a} \right\}$

= $\frac{d}{dx}\left( \frac{x}{2}\sqrt{a^2 - x^2} \right) + \frac{d}{dx}\left( \frac{a^2}{2} \sin^{- 1} \frac{x}{a} \right)$

= $\frac{1}{2}\left[ x\frac{d}{dx}\sqrt{a^2 - x^2} + \sqrt{a^2 - x^2}\frac{d}{dx}\left( x \right) \right] + \frac{a^2}{2} \times \frac{1}{\sqrt{1 - \left( \frac{x}{a} \right)^2}} \times \frac{d}{dx}\left( \frac{x}{a} \right)$

= $\frac{1}{2}\left[ x \times \frac{1}{2\sqrt{a^2 - x^2}}\frac{d}{dx}\left( a^2 - x^2 \right) + \sqrt{a^2 - x^2} \right] + \left[ \frac{a^2}{2} \right] \times \frac{1}{\sqrt{\frac{a^2 - x^2}{a^2}}} \times \left( \frac{1}{a} \right)$

= $\frac{1}{2}\left[ \frac{x\left( - 2x \right)}{2\sqrt{a^2 - x^2}} + \sqrt{a^2 - x^2} \right] + \left( \frac{a^2}{2} \right)\frac{a}{\sqrt{a^2 - x^2}} \times \left( \frac{1}{a} \right)$

= $\frac{1}{2}\left[ \frac{- 2 x^2 + 2\left( a^2 - x^2 \right)}{2\sqrt{a^2 - x^2}} \right] + \frac{a^2}{2\sqrt{a^2 - x^2}}$

= $\frac{1}{2}\left[ \frac{2\left( a^2 - 2 x^2 \right)}{2\sqrt{a^2 - x^2}} \right] + \frac{a^2}{2\sqrt{a^2 - x^2}}$

= $\frac{a^2 - 2 x^2}{2\sqrt{a^2 - x^2}} + \frac{a^2}{2\sqrt{a^2 - x^2}}$

= $\frac{a^2 - 2 x^2 + a^2}{2\sqrt{a^2 - x^2}}$

= $\frac{2 a^2 - 2 x^2}{2\sqrt{a^2 - x^2}}$

= $\frac{2\left( a^2 - x^2 \right)}{2\sqrt{a^2 - x^2}}$

= $\frac{\left( a^2 - x^2 \right)}{\sqrt{a^2 - x^2}}$

= $\sqrt{a^2 - x^2}$

= R.H.S.

Hence proved.