第 12 类 RD Sharma 解决方案 - 第 11 章微分 - 练习 11.2 |设置 2

问题 25. 微分 y = $\log \left( \frac{\sin x}{1 + \cos x} \right)$关于 x。

解决方案：

We have,

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

On differentiating y with respect to x we get,

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

On using chain rule, we have

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

On using quotient rule, we have

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

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

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

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

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

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

问题 26. 微分 y = $\log\sqrt{\frac{1 - \cos x}{1 + \cos x}}$关于 x。

解决方案：

We have,

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

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

As $\log a^b = b\log a$, we get

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

On differentiating y with respect to x we get,

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

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

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

On using chain rule, we have

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

On using quotient rule, we have

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

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

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

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

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

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

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

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

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

问题 27. 微分 y = $\tan \left( e^{\sin x }\right)$关于 x。

解决方案：

We have,

y = $\tan \left( e^{\sin x }\right)$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ \tan\left( e^{\sin x} \right) \right]$

On using chain rule, we have

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

$\frac{d y}{d x} = \sec^2 \left( e^{\sin x} \right) \times e^{\sin x } \times \frac{d}{dx}\left( {\sin x} \right)$

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

问题 28. 微分 y = $\log\left( x + \sqrt{x^2 + 1} \right)$关于 x。

解决方案：

We have,

I = $\log\left( x + \sqrt{x^2 + 1} \right)$

On differentiating y with respect to x we get,

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

On using chain rule, we have

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

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

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

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

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

问题 29. 微分 y = $\frac{e^x \log x}{x^2}$关于 x。

解决方案：

We have,

y = $\frac{e^x \log x}{x^2}$

On differentiating y with respect to x we get,

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

On using quotient rule, we have

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

On using product rule, we have

$\frac{d y}{d x} = \frac{x^2 \left[ \frac{e^x}{x} + e^x \log x \right] - 2x e^x \log x}{x^4}$

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

$\frac{d y}{d x} = \frac{x e^x \left[ 1 + x\log x - 2\log x \right]}{x^4}$

$\frac{d y}{d x} = \frac{x e^x}{x^3}\left[ \frac{1}{x} + \frac{x \log x}{x} - \frac{2\log x}{x} \right]$

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

问题 30. 微分 y = $\log \left( cosec x - \cot x \right)$关于 x。

解决方案：

We have,

y = $\log \left( cosec x - \cot x \right)$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\log \left( cosec x - \cot x \right)$

On using chain rule, we have

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

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

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

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

问题 31. 微分 y = $\frac{e^{2x} + e^{- 2x}}{e^{2x} - e^{- 2x}}$关于 x。

解决方案：

We have,

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

On differentiating y with respect to x we get,

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

On using quotient rule and chain rule, we get

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

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

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

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

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

$\frac{d y}{d x} = \frac{- 8}{\left( e^{2x} - e^{- 2x} \right)^2}$

问题 32. 微分 y = $\log\left( \frac{x^2 + x + 1}{x^2 - x + 1} \right)$关于 x。

解决方案：

We have,

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

On differentiating y with respect to x we get,

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

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

On using quotient rule and chain rule, we get

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

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

$\frac{d y}{d x} = \left( \frac{x^2 - x + 1}{x^2 + x + 1} \right)\left[ \frac{2 x^3 - 2 x^2 + 2x + x^2 - x + 1 - 2 x^3 - 2 x^2 - 2x + x^2 + x + 1}{\left( x^2 - x + 1 \right)^2} \right]$

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

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

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

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

问题 33. 微分 y = $\tan^{- 1} \left( e^x \right)$关于 x。

解决方案：

We have,

y = $\tan^{- 1} \left( e^x \right)$

On differentiating y with respect to x we get,

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

On using chain rule, we have

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

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

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

问题 34. 微分 y = $e^{\sin^{- 1} 2x}$关于 x。

解决方案：

We have,

y = $e^{\sin^{- 1} 2x}$

On differentiating y with respect to x we get,

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

On using chain rule, we have

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

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

$\frac{d y}{d x} = \frac{2 e^{\sin^{- 1} 2x}}{\sqrt{1 - 4 x^2}}$

问题 35. 微分 y = $\sin \left( 2 \sin^{- 1} x \right)$关于 x。

解决方案：

We have,

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

On differentiating y with respect to x we get,

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

On using chain rule, we have

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

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

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

问题 36. 微分 y = $e^{\tan^{- 1} \sqrt{x}}$关于 x。

解决方案：

We have,

y = $e^{\tan^{- 1} \sqrt{x}}$

On differentiating y with respect to x we get,

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

On using chain rule, we have

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

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

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

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

问题 37. 微分 y = $\sqrt{\tan^{- 1} \left( \frac{x}{2} \right)}$关于 x。

解决方案：

We have,

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

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

On differentiating y with respect to x we get,

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

On using chain rule, we have

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

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

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

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

问题 38. 微分 y = $\log\left( \tan^{- 1} x \right)$关于 x。

解决方案：

We have,

y = $\log\left( \tan^{- 1} x \right)$

On differentiating y with respect to x we get,

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

On using chain rule, we have

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

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

问题 39. 微分 y = $\frac{2^x \cos x}{\left( x^2 + 3 \right)^2}$关于 x。

解决方案：

We have,

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

On differentiating y with respect to x we get,

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

On using quotient rule, we have

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

On using product rule and chain rule, we have

$\frac{d y}{d x} = \left[ \frac{\left( x^2 + 3 \right)^2 \left\{ 2^x \frac{d}{dx}\cos x + \cos x\frac{d}{dx} 2^x \right\} - \left( 2^x \cos x \right)2\left( x^2 + 3 \right)\frac{d}{dx}\left( x^2 + 3 \right)}{\left( x^2 + 3 \right)^4} \right]$

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

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

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

问题 40. 微分 y = $x \sin 2x + 5^x + k^k + \left( \tan^2 x \right)^3$关于 x。

解决方案：

We have,

y = $x \sin 2x + 5^x + k^k + \left( \tan^2 x \right)^3$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ x \sin2x + 5^x + k^k + \left( \tan^6 x \right) \right]$

$\frac{d y}{d x} = \frac{d}{dx}\left( x \sin2x \right) + \frac{d}{dx}\left( 5^x \right) + \frac{d}{dx}\left( k^k \right) + \frac{d}{dx}\left( \tan^6 x \right)$

On using product rule and chain rule, we have

$\frac{d y}{d x} = \left[ x\frac{d}{dx}\left( \sin2x \right) + \sin2x\frac{d}{dx}\left( x \right) \right] + 5^x \log_e 5 + 0 + 6 \tan^5 x \times \frac{d}{dx}\left( \tan x \right)$

$\frac{d y}{d x} = \left[ x \cos2x\frac{d}{dx}\left( 2x \right) + \sin2x \right] + 5^x \log_e 5 + 6 \tan^5 x \sec^2 x$

$\frac{d y}{d x} = 2x \cos2x + \sin2x + 5^x \log_e 5 + 6 \tan^5 x \sec^2 x$

问题 41. 微分 y = $\log \left( 3x + 2 \right) - x^2 \log \left( 2x - 1 \right)$关于 x。

解决方案：

We have,

y = $\log \left( 3x + 2 \right) - x^2 \log \left( 2x - 1 \right)$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ \log\left( 3x + 2 \right) - x^2 \log\left( 2x - 1 \right) \right]$

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

On using product rule and chain rule, we have

$\frac{d y}{d x} = \frac{1}{\left( 3x + 2 \right)}\frac{d}{dx}\left( 3x + 2 \right) - \left[ x^2 \frac{d}{dx}\log\left( 2x - 1 \right) + \log\left( 2x - 1 \right)\frac{d}{dx}\left( x^2 \right) \right]$

$\frac{d y}{d x} = \frac{3}{3x + 2} - \frac{2 x^2}{\left( 2x - 1 \right)} - 2x \log\left( 2x - 1 \right)$

问题 42. 微分 y = $\frac{3 x^2 \sin x}{\sqrt{7 - x^2}}$关于 x。

解决方案：

We have,

y = $\frac{3 x^2 \sin x}{\sqrt{7 - x^2}}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left\{ \frac{3 x^2 sinx}{\left( 7 - x^2 \right)^\frac{1}{2}} \right\}$

On using quotient rule, chain rule and product rule we get,

$\frac{d y}{d x} = \frac{\left( 7 - x^2 \right)^\frac{1}{2} \times \frac{d}{dx}\left( 3 x^2 \sin x \right) - \left( 3 x^2 \sin x \right)\frac{d}{dx} \left( 7 - x^2 \right)^\frac{1}{2}}{\left[ \left( 7 - x^2 \right)^\frac{1}{2} \right]^2} \left[ \text{} \right]$

$\frac{d y}{d x} = \left[ \frac{\left( 7 - x^2 \right)^\frac{1}{2} \times 3\left[ x^2 \frac{d}{dx}\left( \sin x \right) + \sin x\frac{d}{dx}\left( x^2 \right) \right] - 3 x^2 \sin x \times \frac{1}{2}\left( 7 - x^2 \right) \times \frac{d}{dx}\left( 7 - x^2 \right)}{\left( 7 - x^2 \right)} \right]$

$\frac{d y}{d x} = \left[ \frac{\left( 7 - x^2 \right)^\frac{1}{2} 3\left( x^2 \cos x + 2x \sin x \right) - 3 x^2 \sin x \times \frac{1}{2} \left( 7 - x^2 \right)^\frac{- 1}{2} \left( - 2x \right)}{\left( 7 - x^2 \right)} \right]$

$\frac{d y}{d x} = \left[ \frac{\left( 7 - x^2 \right)^\frac{1}{2} \times 3\left( x^2 \cos x + 2x \sin x \right)}{\left( 7 - x^2 \right)} + \frac{3 x^3 \sin x \left( 7 - x^2 \right)^\frac{- 1}{2}}{\left( 7 - x^2 \right)} \right]$

$\frac{d y}{d x} = \left[ \frac{6x \sin x + 3 x^2 \cos x}{\sqrt{\left( 7 - x^2 \right)}} + \frac{3 x^3 \sin x}{\left( 7 - x^2 \right)^\frac{3}{2}} \right]$

问题 43. 微分 y = $\sin^2 \left\{ \log \left( 2x + 3 \right) \right\}$关于 x。

解决方案：

We have,

y = $\sin^2 \left\{ \log \left( 2x + 3 \right) \right\}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ \sin^2 \left\{ \log\left( 2x + 3 \right) \right\} \right]$

On using chain rule, we get

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

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

As 2 sin A cos A = sin 2A, we get

$\frac{d y}{d x} = \sin\left\{ 2\log\left( 2x + 3 \right) \right\} \times \frac{1}{\left( 2x + 3 \right)}\frac{d}{dx}\left( 2x + 3 \right)$

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

问题 44. 微分 y = $e^x \log \sin 2x$关于 x。

解决方案：

We have,

y = $e^x \log \sin 2x$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ e^x \log \sin2x \right]$

On using product rule and chain rule, we have

$\frac{d y}{d x} = e^x \frac{d}{dx}\left( \log \sin2x \right) + \left( \log \sin2x \right)\frac{d}{dx}\left( e^x \right)$

$\frac{d y}{d x} = e^x \frac{1}{\sin2x}\frac{d}{dx}\left( \sin2x \right) + \log \sin2x\left( e^x \right)$

$\frac{d y}{d x} = \frac{e^x}{\sin2x}\cos2x \frac{d}{dx}\left( 2x \right) + e^x \log \sin2x$

$\frac{d y}{d x} = \frac{2\cos2x e^x}{\sin2x} + e^x \log \sin2x$

$\frac{d y}{d x} = 2 e^x \cot2x + e^x \log \sin2x$

问题 45. 微分 y = $\frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} - \sqrt{x^2 - 1}}$关于 x。

解决方案：

We have,

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

On rationalizing we get,

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

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

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

y = $\frac{x^2 + 1 + x^2 - 1 + 2\sqrt{x^4 - 1}}{2}$

y = $\frac{2 x^2 + 2\sqrt{x^4 - 1}}{2}$

y = $x^2 + \sqrt{x^4 - 1}$

On differentiating y with respect to x we get,

$\frac{dy}{dx} = \frac{d}{dx}\left( x^2 + \sqrt{x^4 - 1} \right)$

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

$\frac{d y}{d x} = 2x + \frac{1}{2\sqrt{x^4 - 1}} \times \left( 4 x^3 \right)$

$\frac{d y}{d x} = 2x + \frac{2 x^3}{\sqrt{x^4 - 1}}$

问题 46. 微分 y = $\log [x+2+\sqrt{x^2+4x+1}]$关于 x。

解决方案：

We have,

y = $\log [x+2+\sqrt{x^2+4x+1}]$

On differentiating y with respect to x we get,

$\frac{d y}{d x}=\frac{d}{dx}\log[x+2+\sqrt{x^2+4x+1}]$

On using chain rule, we have

$\frac{d y}{d x}=\frac{1}{([x+2+sqrt(x^4+4x+1)])}\frac{d}{dx}[x+2+(x^2+4x+1)^{\frac{1}{2}}]$

$\frac{d y}{d x}=\frac{1}{x+2+sqrt(x^4+4x+1)}[1+0+\frac{1}{2}(x^2+4x+1)^{-1/2}\frac{d}{dx}(x^2+4x+1)]$

$\frac{d y}{d x}=\frac{1+\frac{2x+4}{2\sqrt{x^2+4x+1}}}{[x+2+\sqrt{x^4+4x+1}]}$

$\frac{d y}{d x}=\frac{\sqrt{x^4+4x+1}+x+2}{[x+2+\sqrt{x^4+4x+1}]\sqrt{x^4+4x+1}}$

$\frac{d y}{d x}=\frac{1}{\sqrt{x^2+4x+1}}$

问题 47. 微分 y = $\left( \sin^{- 1} x^4 \right)^4$关于 x。

解决方案：

We have,

y = $\left( \sin^{- 1} x^4 \right)^4$

On differentiating y with respect to x we get,

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

On using chain rule, we have

$\frac{d y}{d x} = 4 \left( \sin^{- 1} x^4 \right)^3 \frac{d}{dx}\left( \sin^{- 1} x^4 \right)$

On using chain rule again, we have

$\frac{d y}{d x} = 4 \left( \sin^{- 1} x^4 \right)^3 \frac{1}{\sqrt{1 - \left( x^4 \right)^2}}\frac{d}{dx}\left( x^4 \right)$

$\frac{d y}{d x} = 4 \left( \sin^{- 1} x^4 \right)^3 \frac{4 x^3}{\sqrt{1 - x^8}}$

$\frac{d y}{d x} = \frac{16 x^3 \left( \sin^{- 1} x^4 \right)^3}{\sqrt{1 - x^8}}$

问题 48. 微分 y = $\sin^{- 1} \left( \frac{x}{\sqrt{x^2 + a^2}} \right)$关于 x。

解决方案：

We have,

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

On differentiating y with respect to x we get,

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

On using chain rule and quotient rule, we get

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

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

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

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

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

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

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

问题 49. 微分 y = $\frac{e^x \sin x}{\left( x^2 + 2 \right)^3}$关于 x。

解决方案：

We have,

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

On differentiating y with respect to x we get,

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

On using quotient rule, we get

$\frac{d y}{d x} = \frac{\left( x^2 + 2 \right)^3 \frac{d}{dx}\left( e^x \sin x \right) - e^x \sin x\frac{d}{dx} \left( x^2 + 2 \right)^3}{\left[ \left( x^2 + 2 \right)^3 \right]^2}$

On using product rule, we get

$\frac{d y}{d x} = \frac{\left( x^2 + 2 \right)^3 \left[ e^x \cos x + \sin x e^x \right] - e^x \sin x 3 \left( x^2 + 2 \right)^2 \left( 2x \right)}{\left( x^2 + 2 \right)^6}$

$\frac{d y}{d x} = \frac{\left( x^2 + 2 \right)^3 \left[ e^x \cos x + e^x \sin x \right] - 6x e^x \sin x \left( x^2 + 2 \right)^2}{\left( x^2 + 2 \right)^6}$

$\frac{d y}{d x} = \frac{\left( x^2 + 2 \right)^2 \left[ \left( x^2 + 2 \right)\left( e^x \cos x + e^x \sin x \right) - 6x e^x \sin x \right]}{\left( x^2 + 2 \right)^6}$

$\frac{d y}{d x} = \frac{\left( x^2 + 2 \right)\left( e^x \cos x + e^x \sin x \right) - 6x e^x \sin x}{\left( x^2 + 2 \right)^4}$

$\frac{d y}{d x} = \frac{e^x \sin x + e^x \cos x}{\left( x^2 + 2 \right)^3} - \frac{6x e^x \sin x}{\left( x^2 + 2 \right)^4}$