第 12 类 NCERT 解决方案 – 数学第一部分 – 第 5 章连续性和可微性 – 练习 5.5 | 2套

第5章连续性和可微性——练习5.5 |设置 1

问题 11. 关于 x 对函数进行微分。

(x cos x) ^x + (x sin x) ^1/x

解决方案：

Given: (x cos x)^x+ (x sin x)^1/x

Let us considered y = u + v

Where, u = (x cos x)^xand v = (x sin x)^1/x

So, dy/dx = du/dx + dv/dx ………(1)

So first we take u = (x cos x)^x

On taking log on both sides, we get

log u = log(x cos x)^x

log u = xlog(x cos x)

Now, on differentiating w.r.t x, we get

$\frac{1}{u}\frac{du}{dx}=x\frac{d}{dx}(\log x+\log(\cos x))+\log x+\log \cos x$

$\frac{1}{u}\frac{du}{dx}=x(\frac{1}{x}+\frac{1}{\cos x}\frac{d}{dx}\cos x)+\log x+\log\cos x$

$\frac{du}{dx}=u(x(\frac{1}{x}+\frac{-\sin x}{\cos x})+\log x+\log(\cos x))$

$\frac{du}{dx}=(x\cos x)^x(1-x\tan x+\log x+\log(\cos x))$ ………(2)

Now we take u =(x sin x)^1/x

On taking log on both sides, we get

log v = log (x sin x)^1/x

log v = 1/x log (x sin x)

log v = 1/x(log x + log sin x)

Now, on differentiating w.r.t x, we get

$\frac{1}{v}\frac{dv}{dx}=\frac{1}{x}\frac{d}{dx}(\log x+\log(\sin x)+\frac{d}{dx}(\frac{1}{x}).(\log x+\log(\sin x)))$

$\frac{1}{v}.\frac{dv}{dx}=\frac{1}{x}(\frac{1}{x}+\frac{1}{\sin x}.\frac{d}{dx}\sin x)+(\frac{-1}{x^2})(\log x+\log(\sin x))$

$\frac{dv}{dx}=v(\frac{1}{x}(\frac{1}{x}+\frac{\cos x}{\sin x})\frac{-1}{x^2}(\log x+\log(\sin x)))$

$\frac{dv}{dx}=(x\sin x)^{1/2}.[(\frac{1}{x^2}+\frac{\cot x}{x})-\frac{\log x}{x^2}-\frac{\log(\sin x)}{x^2}]$ ………(3)

Now put all the values from eq(2) and (3) into eq(1)

$\frac{dy}{dx}=(x\cos)^x(1-x\tan x+\log x+\log(\cos x))+(x\sin x)^{\frac{1}{x}}.[\frac{xcotx+1-log(xsinx)}{x^2}]$

编程需要懂一点英语

找出问题 12 到 15 中给出的函数的dy/dx

问题 12. x ^y + y ^x = 1

解决方案：

Given: x^y+ y^x= 1

Let us considered

u = x^y and v = y^x

So,

$\frac{du}{dx}+\frac{dv}{dx}=0$ ………(1)

So first we take u = x^y

On taking log on both sides, we get

log u = log(x^y)

log u = y log x

Now, on differentiating w.r.t x, we get

$\frac{1}{u}.\frac{du}{dx}=y.\frac{d}{dx}\log x+\frac{dy}{dx}.\log x$

$\frac{1}{u}\frac{du}{dx}=\frac{y}{x}+\frac{dy}{dx}\log x$

$\frac{du}{dx}=x^4(\frac{y}{x}+\frac{dy}{dx}\log x)$ ………(2)

Now we take v = y^x

On taking log on both sides, we get

log v = log(y)^x

log v = x log y

Now, on differentiating w.r.t x, we get

$\frac{1}{v}.\frac{dv}{dx}=x\frac{d}{dx}(\log x)+\log y\frac{d}{dx}x$

$\frac{dv}{dx}=v(x.\frac{1}{y}.\frac{dy}{dx}+\log y)$

$\frac{dv}{dx}=y^x(\frac{x}{y}\frac{dy}{dx}+\log x)$ ………(3)

Now put all the values from eq(2) and (3) into eq(1)

$x^y(\frac{y}{x}+\frac{dy}{dx}\log x)+y^x(\frac{x}{y}\frac{dy}{dx}+\log y)=0$

$(x^y.\log x+xy^{x-1})\frac{dy}{dx}=-(yx^{y-1}+y^x\log y)$

$\frac{dy}{dx}=\frac{-yx^{y-1}+y^x\log y}{x^y\log x+xy^{x-1}}$

编程需要懂一点英语

问题 13. y ^x = x ^y

解决方案：

Given: y^x= x^y

On taking log on both sides, we get

log(yx) = log(x^y)

xlog y = y log x

Now, on differentiating w.r.t x, we get

$x\frac{dy}{dx}(\log y)+\log y(\frac{d}{dx}x)=y\frac{d}{dx}\log x+\log x\frac{d}{dx}y$

$x.\frac{d}{dx}.y+\log y.1=y.\frac{1}{x}+\log x\frac{dy}{dx}$

$\frac{x}{y}\frac{dy}{dx}+\log y=\frac{y}{x}+\log x\frac{dy}{dx}$

$(\frac{x}{y}-\log x)\frac{dy}{dx}=(\frac{y}{x}-\log y)$

$\frac{dy}{dx}=\frac{\frac{y}{x}-\log y}{\frac{x}{y}-\log x}$

$\frac{dy}{dx}=\frac{y}{x}(\frac{y-x\log y}{x-y\log x})$

编程需要懂一点英语

问题 14. (cos x) ^y = (cos y) ^x

解决方案：

Given: (cos x)^y= (cos y)^x

On taking log on both sides, we get

y log(cos x) = x log (cos y)

Now, on differentiating w.r.t x, we get

$y\frac{d}{dx}\log(\cos x)+\log(\cos x).\frac{dy}{dx}=x\frac{d}{dx}\log (\cos y)+\log(\cos y)\frac{dx}{dx}$

$y\frac{1}{\cos x}\frac{d}{dx}\cos x+\log(\cos x)\frac{dy}{dx}=x\frac{1}{\cos y}\frac{d}{dx}\cos y+\log(\cos y).1$

$\frac{y}{\cos x}.-\sin x+\log(\cos x).\frac{dy}{dx}=\frac{x}{\cos y}.(-\sin y).\frac{dy}{dx}+\log(\cos y)$

$(\log(\cos x)+x\tan y)\frac{dy}{dx}=\log(\cos y)+y \tan x$

$\frac{dy}{dx}=\frac{\log(\cos y)+y\tan x}{\log(\cos x)+x\tan y}$

编程需要懂一点英语

问题 15. xy = e ^{(x – y)}

解决方案：

Given: xy = e^{(x – y)}

On taking log on both sides, we get

log(xy) = log e^{x – y}

log x + log y = x – y

Now, on differentiating w.r.t x, we get

$\frac{d}{dx}\log x+\frac{d}{dx}\log y=\frac{dx}{dx}-\frac{dy}{dx}$

$\frac{1}{x}.+\frac{1}{y}.\frac{dy}{dx}=1-\frac{dy}{dx}$

$(\frac{1}{y}+1)\frac{dy}{dx}=(1-\frac{1}{x})$

$\frac{dy}{dx}=\frac{(1-\frac{1}{x})}{(1+\frac{1}{y})}$

$\frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}$

编程需要懂一点英语

问题 16. 找出由 f(x) = (x + 1)(x + x ² )(1 + x ⁴ )(1 + x ⁸ ) 给出的函数的导数，从而找到 f'(1)。

解决方案：

Given: f(x) = (x + 1)(x + x²)(1 + x⁴)(1 + x⁸)

Find: f'(1)

On taking log on both sides, we get

log(f(x)) = log(1 + x) + log(1 + x²) + log(1 + x⁴) + log(1 + x⁸)

Now, on differentiating w.r.t x, we get

$\frac{1}{f(x)}.\frac{d}{dx}f{x}=\frac{1}{1+x}\frac{d}{dx}(1+x)+\frac{1}{1+x^2}(1+x)^2+\frac{1}{1+x^4}.\frac{d}{dx}(1+x^4)+\frac{1}{1+x^8}\frac{d}{dx}(1+x^8)\frac{f'(x)}{f(x)}=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}$

$f'(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)(\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4x^3}{1+x^2}+\frac{8x^7}{1+x^8})$

∴ f'(1) = 2.2.2.2. $(\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2})$

$f'(1)=16.(\frac{15}{2})$

f'(1) = 120

编程需要懂一点英语

问题 17. 用下面提到的三种方法区分 (x ⁵ – 5x + 8)(x ³ + 7x + 9)

(i) 通过使用乘积规则

(ii) 通过展开乘积得到单个多项式

(iii) 通过对数微分。

他们都给出相同的答案吗？

解决方案：

(i) By using product rule

$\frac{d}{dx}(u.v)=v\frac{du}{dv}+u\frac{dv}{dx}$

$\frac{dy}{dx}=(x^2-5x+8)\frac{d}{dx}(x^3+7x+9)+(x^3+7x+9).\frac{d}{dx}(x^2-5x+8)$

dy/dx = (3x⁴– 15x³+ 24x²+ 7x²– 35x + 56) + (2x⁴+ 14x²+ 18x – 5x³– 35x – 45)

dy/dx = 5x⁴– 20x³+ 45x²– 52x + 11

(ii) By expansion

y = (x²– 5x + 8)(x³+ 7x + 9)

y = x⁵+ 7x³+ 9x²– 5x⁴– 35x²– 45x + 8x³+ 56x + 72

y = x⁵– 5x⁴+ 15x³– 26x²+ 11x + 72

dy/dx = 5x⁴– 20x³+ 45x²– 52x + 11

(iii) By logarithmic expansion

Taking log on both sides

log y = log(x²– 5x + 8) + log(x³+ 7x + 9)

Now on differentiating w.r.t. x, we get

$\frac{1}{y}.\frac{dy}{dx}=\frac{1}{x^2-5x+8}.\frac{d}{dx}(x^2-5x+8)+\frac{1}{x^3+7x+9}\frac{d}{dx}(x^3+7x+9)$

$\frac{1}{y}.\frac{dy}{dx}=\frac{2x-5}{x^2-5x+8}+\frac{3x^2+7}{x^3+7x+9}$

$\frac{1}{(x^2-5x+8)(x^3+7x+9)}\frac{dy}{dx}=\frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)}$

dy/dx = 2x⁴+ 14x²+ 18x – 5x³– 35x – 45 + 3x⁴– 15x³+ 24x²+ 7x²– 35x + 56

dy/dx = 5x⁴– 20x³+ 45x²– 52x + 11

Answer is always same what-so-ever method we use.

编程需要懂一点英语

问题 18. 如果 u、v 和 w 是 x 的函数，则证明

$\frac{d}{dx}(u.v.w)=\frac{du}{dx}v.w+u.\frac{dv}{dx}.w+u.v.\frac{dw}{dx}$

解决方案：

Let y = u.v.w.

Method 1: Using product Rule

$\frac{dy}{dx}=u\frac{d}{dx}(v.w)+v.w\frac{d}{dx}u$

$\frac{dy}{dx}=u.[v.\frac{dw}{dx}+w\frac{du}{dx}]+v.w.\frac{du}{dx}$

$\frac{dy}{dx}=u.v.\frac{dw}{dx}+u.w.\frac{dv}{dx}+v.w\frac{du}{dx}$

Method 2: Using logarithmic differentiation

Taking log on both sides

log y = log u + log v + log w

Now, Differentiating w.r.t. x

$\frac{1}{y}\frac{dy}{dx}=\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}.\frac{dw}{dx}$

$\frac{dy}{dx}=(u.v.w)(\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx})$

$\frac{dy}{dx}=v.w\frac{du}{dx}+uw\frac{dv}{dx}+uv\frac{dw}{dx}$

编程需要懂一点英语

第 12 类 NCERT 解决方案 – 数学第一部分 – 第 5 章连续性和可微性 – 练习 5.5 | 2套

第5章连续性和可微性——练习5.5 |设置 1

问题 11. 关于 x 对函数进行微分。

(x cos x) x + (x sin x) 1/x

找出问题 12 到 15 中给出的函数的dy/dx

问题 12. x y + y x = 1

问题 13. y x = x y

问题 14. (cos x) y = (cos y) x

问题 15. xy = e (x – y)

问题 16. 找出由 f(x) = (x + 1)(x + x 2 )(1 + x 4 )(1 + x 8 ) 给出的函数的导数，从而找到 f'(1)。

问题 17. 用下面提到的三种方法区分 (x 5 – 5x + 8)(x 3 + 7x + 9)

(i) 通过使用乘积规则

(ii) 通过展开乘积得到单个多项式

(iii) 通过对数微分。

他们都给出相同的答案吗？

问题 18. 如果 u、v 和 w 是 x 的函数，则证明

(x cos x) ^x + (x sin x) ^1/x

问题 12. x ^y + y ^x = 1

问题 13. y ^x = x ^y

问题 14. (cos x) ^y = (cos y) ^x

问题 15. xy = e ^{(x – y)}

问题 16. 找出由 f(x) = (x + 1)(x + x ² )(1 + x ⁴ )(1 + x ⁸ ) 给出的函数的导数，从而找到 f'(1)。

问题 17. 用下面提到的三种方法区分 (x ⁵ – 5x + 8)(x ³ + 7x + 9)