第 11 类 RD Sharma 解决方案 - 第 30 章衍生品 - 练习 30.4 |设置 3

问题 21. 将 (2x ² – 3) sin x 与 x 微分。

解决方案：

We have,

=> y = (2x² – 3) sin x

On differentiating both sides, we get,

$\frac{dy}{dx}=\frac{d}{dx}[(2x^2 - 3) sin x]$

On using product rule we get,

= $sinx\frac{d}{dx}(2x^2 - 3)+(2x^2 - 3)\frac{d}{dx}(sinx)$

= $sinx(4x)+(2x^2-3)cosx$

= $4xsinx+2x^2cosx-3cosx$

问题 22. 区分$x^5(3-6x^{-9})$关于 x。

解决方案：

We have,

=> y = $x^5(3-6x^{-9})$

On differentiating both sides, we get,

$\frac{dy}{dx}=\frac{d}{dx}[x^5(3-6x^{-9})]$

On using product rule we get,

= $(3-6x^{-9})\frac{d}{dx}(x^5)+(x^5)\frac{d}{dx}(3-6x^{-9})$

= $(3-6x^{-9})(5x^4)+(x^5)[-(6)(-9)x^{-10}]$

= $(3-6x^{-9})(5x^4)+(x^5)(54x^{-10})$

= $15x^4-30x^5+54x^{-5}$

问题 23. 区分$x^{-4}(3-4x^{-5})$关于 x。

解决方案：

We have, 

=> y = $x^{-4}(3-4x^{-5})$

On differentiating both sides, we get,

$\frac{dy}{dx}=\frac{d}{dx}[x^{-4}(3-4x^{-5})]$

On using product rule we get,

= $(3-4x^{-5})\frac{d}{dx}(x^{-4})+x^{-4}\frac{d}{dx}(3-4x^{-5})$

= $(3-4x^{-5})(-4x^{-5})+x^{-4}[-(4)(-5)x^{-6}]$

= $(3-4x^{-5})(-4x^{-5})+x^{-4}[20x^{-6}]$

= $-12x^{-5}+16x^{-10}+20x^{-10}$

= $-12x^{-5}+36x^{-10}$

= $-12x^{-5}(1+3x^{-5})$

问题 24. 区分$x^{-3}(5+3x)$关于 x。

解决方案：

We have,

=> y = $x^{-3}(5+3x)$

On differentiating both sides, we get,

$\frac{dy}{dx}=\frac{d}{dx}[x^{-3}(5+3x)]$

On using product rule we get, 

= $(5+3x)\frac{d}{dx}(x^{-3})+x^{-3}\frac{d}{dx}(5+3x)$

= $(5+3x)(-3x^{-4})+x^{-3}(3)$

= $-15x^{-4}-9x^{-3}+3x^{-3}$

= $-15x^{-4}-6x^{-3}$

= $-3x^{-3}(5x^{-1}-2)$

问题 25. 区分$\frac{ax+b}{cx+d}$关于 x。

解决方案：

We have, 

=> y = $\frac{ax+b}{cx+d}$

On differentiating both sides, we get,

$\frac{dy}{dx}=\frac{d}{dx}[\frac{ax+b}{cx+d}]$

On using product rule we get, 

= $(\frac{1}{cx+d})\frac{d}{dx}(ax+b)+(ax+b)\frac{d}{dx}(\frac{1}{cx+d})$

= $(\frac{1}{cx+d})(a)+(ax+b)\left[\frac{-1}{(cx+d)^{2}}×c\right]$

= $(\frac{a}{cx+d})+(ax+b)\left[\frac{-c}{(cx+d)^{2}}\right]$

= $(\frac{a}{cx+d})-\frac{c(ax+b)}{(cx+d)^{2}}$

= $\frac{a(cx+d)-c(ax+b)}{(cx+d)^2}$

= $\frac{acx+ad-acx-bc}{(cx+d)^2}$

= $\frac{ad-bc}{(cx+d)^2}$

问题 26. 将 (ax + b) ⁿ (cx + d) ^m与 x 微分。

解决方案：

We have,  

=> y = (ax + b)ⁿ (cx + d)^m

On differentiating both sides, we get,

$\frac{dy}{dx}=\frac{d}{dx}[(ax + b)^n (cx + d)^m]$

On using product rule we get, 

= $(cx + d)^m\frac{d}{dx}[(ax + b)^n]+(ax + b)^n\frac{d}{dx}[(cx + d)^m]$

= $(cx + d)^m[na(ax + b)^{n-1}]+(ax + b)^n[mc(cx + d)^{m-1}]$

= $na(cx + d)^m(ax + b)^{n-1}+mc(ax + b)^n(cx + d)^{m-1}$

= $(cx + d)^{m-1}(ax + b)^{n-1}[na(cx+d)+mc(ax + b)]$

问题 27. 以两种方式进行微分，使用乘积规则，否则使用函数(1 + 2 tan x) (5 + 4 cos x)。验证答案是否相同。

解决方案：

We have, 

=> y = (1 + 2 tan x) (5 + 4 cos x)

On using product rule we get, 

$\frac{d}{dx}[(1 + 2 tan x) (5 + 4 cos x)]=(5 + 4 cos x)\frac{d}{dx}(1+2tanx)+(1+2tanx)\frac{d}{dx}(5+4cosx)$

= $(5+4cosx)(2sec^2x)+(1+2tanx)(-4sinx)$

= 10 sec² x + 8 cos x sec² x − 4 sin x − 8 sin x tan x

=  $10sec^2x+8cosx(\frac{1}{cos^2x})−4sinx−8sinx(\frac{sinx}{cosx})$

= $10sec^2x+\frac{8}{cosx}−4sinx−\frac{8sin^2x}{cosx}$

= $10sec^2x+\frac{8(1-sin^2x)}{cosx}−4sinx$

= $10sec^2x+\frac{8cos^2x}{cosx}−4sinx$

= 10 sec² x + 8 cos x − 4 sin x 

By using an alternate method, we have,

$\frac{d}{dx}[(1 + 2 tan x) (5 + 4 cos x)]=\frac{d}{dx}(5+4cosx+10tanx+8tanxcosx)$

= $\frac{d}{dx}(5+4cosx+10tanx+8(\frac{sinx}{cosx})cosx)$

= $\frac{d}{dx}(5+4cosx+10tanx+8sinx)$

On using chain rule, we get,

= 0 − 4 sin x + 10 sec² x + 8 cos x

= 10 sec² x + 8 cos x − 4 sin x 

Hence proved.

问题 28. 通过乘积规则和其他方法区分以下每个功能，并验证两种方法的答案是否相同。

(一) (3x ² + 2) ²

解决方案：

We have, 

=> y = (3x² + 2)²

On using product rule we get,

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

= $(3x^2+2)(6x)+(3x^2+2)(6x)$

= 12x (3x² + 2)

= 36 x³ + 24x

By using an alternate method, we have,

$\frac{d}{dx}[(3x^2+2)^2]=\frac{d}{dx}[9x^4+4+12x^2]$

On using chain rule, we get,

= 36 x³ + 0 + 24 x

= 36 x³ + 24x

Hence proved.

(ii) (x + 2)(x + 3)

解决方案：

We have,

=> y = (x + 2)(x + 3)

On using product rule we get,

$\frac{d}{dx}[(x + 2)(x + 3)]=(x+3)\frac{d}{dx}(x+2)+(x+2)\frac{d}{dx}(x+3)$

= (x+3)(1)+(x+2)(1)

= x + 3 + x + 2

= 2x + 5

By using an alternate method, we have,

$\frac{d}{dx}[(x + 2)(x + 3)]=\frac{d}{dx}[x^2+5x+6]$

On using chain rule, we get,

= 2x + 5

Hence proved.

(iii) (3 sec x − 4 cosec x) (−2 sin x + 5 cos x)

解决方案：

We have, 

=> y = (3 sec x − 4 cosec x) (−2 sin x + 5 cos x)

On using product rule we get,

$\frac{d}{dx}[(3secx−4cosecx)(−2sinx+5cosx)]=(−2sinx+5cosx)\frac{d}{dx}(3secx−4cosecx)+(3secx−4cosecx)\frac{d}{dx}(−2sinx+5cosx)$

= (−2 sin x + 5 cos x) (3 sec x tan x + 4 cot x cosec x)+ (3 sec x − 4 cosec x) (−2 cos x − 5 sin x)

= −6 sin x sec x tan x − 8 sin x cot x cosec x + 15 cos x sec x tan x  + 20 cos x cot x cosec x − 6 sec x cos x − 15 sec x sin x + 8 cosec x cos x + 20 cosec x sin x

= −6 tan² x − 8 cot x + 15 tan x + 20 cot² x − 6 − 15 tan x + 8 cot x + 20

= − 6 − 6 tan² x + 20 cot² x + 20

= −6 (1 + tan² x) + 20 (cot² x + 1)

= −6 sec² x + 20 cosec² x

By using an alternate method, we have,

$\frac{d}{dx}[(3secx−4cosecx)(−2sinx+5cosx)]=\frac{d}{dx}(−6secxsinx+15secxcosx+6cosecxsinx-20cosecxcosx)$

= $\frac{d}{dx}(−6tanx+15+6-20cotx)$

= $\frac{d}{dx}(−6tanx-20cotx+21)$

On using chain rule, we get,

= −6 sec² x − (−20 cosec² x)

= −6 sec² x + 20 cosec² x

Hence proved.