11类NCERT解决方案–第13章限制和导数–练习13.2

问题1.在x = 10处找到x ² – 2的导数。

解决方案：

f(x) = x² – 2

f(x+h) = (x+h)² – 2

From the first principle,

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

When, x = 10

$f'(10) = \lim_{h \to 0} (\frac{f(10+h)-f(10)}{h})\\ f'(10) = \lim_{h \to 0} (\frac{((10+h)^2-2) - (10^2-2)}{h})\\ f'(10) = \lim_{h \to 0} (\frac{((10^2+2(10)h+h^2)-2) - (10^2-2)}{h})\\ f'(10) = \lim_{h \to 0} (\frac{(10^2+2(10)h+h^2-2 - 10^2+2)}{h})\\ f'(10) = \lim_{h \to 0} (\frac{(20h+h^2)}{h})\\ f'(10) = \lim_{h \to 0} (20 + h)$

f'(10) = 20 + 0

f'(10) = 20

问题2.在x = 1处找到x的导数。

解决方案：

f(x) = x

f(x+h) = x+h

From the first principle,

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

When, x = 1

$f'(1) = \lim_{h \to 0} (\frac{f(1+h)-f(1)}{h})\\ f'(1) = \lim_{h \to 0} (\frac{(1+h) - (1)}{h})\\ f'(1) = \lim_{h \to 0} (\frac{(1+h - 1)}{h})\\ f'(1) = \lim_{h \to 0} (\frac{h}{h})\\ f'(1) = \lim_{h \to 0} (1)$

f'(1) = 1

问题3.在x = 100处找到99x的导数。

解决方案：

f(x) = 99x

f(x+h) = 99(x+h)

From the first principle,

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

When, x = 10

$f'(100) = \lim_{h \to 0} (\frac{f(100+h)-f(100)}{h})\\ f'(100) = \lim_{h \to 0} (\frac{((99(100+h) - (99(100))}{h})\\ f'(100) = \lim_{h \to 0} (\frac{(9900+99h - 9900)}{h})\\ f'(100) = \lim_{h \to 0} (\frac{99h}{h})\\ f'(100) = \lim_{h \to 0} (99)$

f'(100) = 99

问题4.从第一原理中找到以下函数的派生词。

(i)x ³ − 27

解决方案：

f(x) = x³ – 27

f(x+h) = (x+h)³ – 27

From the first principle,

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} \frac{(x+h)^3 - 27-(x^3 - 27)}{h}\\ f'(x) = \lim_{h \to 0} \frac{(x+h)^3 -x^3}{h}\\ f'(x) = \lim_{h \to 0} \frac{x^3+h^3+3xh(x+h)-x^3}{h}\\ f'(x) = \lim_{h \to 0} \frac{h^3+3xh(x+h)}{h}\\ f'(x) = \lim_{h \to 0} (h^2+3x(x+h))$

f'(x) = 0²+3x(x+0)

f'(x) = 3x²

(ii)(x-1)(x-2)

解决方案：

f(x) = (x-1) (x-2) = x² – 3x + 2

f(x) = (x+h)² – 3(x+h) + 2

From the first principle,

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - 3(x+h) + 2-(x^2 - 3x + 2)}{h}\\ f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - 3(x+h) + 2-x^2 + 3x - 2)}{h}\\ f'(x) = \lim_{h \to 0} \frac{(x^2+2xh+h^2 - 3x - 3h + 2-x^2 + 3x - 2)}{h}\\ f'(x) = \lim_{h \to 0} \frac{(2xh+h^2 - 3h)}{h}\\ f'(x) = \lim_{h \to 0} (2x+h - 3)$

f'(x) = 2x+0 – 3

f'(x) = 2x – 3

(iii) $\frac{1}{x^2}$

解决方案：

$f(x) = \frac{1}{x^2}\\ f(x) = \frac{1}{(x+h)^2}$

From the first principle,

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2}-(\frac{1}{x^2})}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{x^2-(x^2+2xh+h^2)}{x^2(x+h)^2}}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{x^2-x^2-2xh-h^2)}{x^2(x+h)^2}}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{-2xh-h^2)}{x^2(x+h)^2}}{h}\\ f'(x) = \lim_{h \to 0} \frac{-2x-h)}{x^2(x+h)^2}\\ f'(x) = \frac{-2x-0)}{x^2(x+0)^2}\\ f'(x) = \frac{-2x}{x^2(x)^2}\\ f'(x) = \frac{-2}{x^3}$

(iv) $\frac{x+1}{x-1}$

解决方案：

$f(x) = \frac{x+1}{x-1}\\ f(x) = \frac{(x+h)+1}{(x+h)-1}$

From the first principle,

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{(x+h)+1}{(x+h)-1}-(\frac{x+1}{x-1})}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{(x+h+1)(x-1)-(x+h-1)(x+1)}{(x+h-1)(x-1)}}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{(x^2+hx+x)-(x+h+1)-[(x^2+hx-x)+(x+h-1)]}{(x+h-1)(x-1)}}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{(x^2+hx+x-x-h-1)-(x^2+hx-x+x+h-1)}{(x+h-1)(x-1)}}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{(x^2+hx+x-x-h-1-x^2-hx+x-x-h+1)}{(x+h-1)(x-1)}}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{(-h-h)}{(x+h-1)(x-1)}}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{-2h}{(x+h-1)(x-1)}}{h}\\ f'(x) = \lim_{h \to 0} \frac{-2h}{h(x+h-1)(x-1)}\\ f'(x) = \lim_{h \to 0} \frac{-2}{(x+h-1)(x-1)}\\ f'(x) = \frac{-2}{(x-1)(x-1)}\\ f'(x) = \frac{-2}{(x-1)^2}\\$

问题5.对于函数

f(x)= $\frac{x^{100}}{100} + \frac{x^{99}}{99} + ......... + \frac{x^2}{2} + x + 1.$

证明f’(1)= 100 f’(0)

解决方案：

Given,

$f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + ......... + \frac{x^2}{2} + x + 1.$

By using this, taking derivative both sides

$f'(x) = \frac{d}{dx}(\frac{x^{100}}{100}) + \frac{d}{dx}(\frac{x^{99}}{99}) + ......... + \frac{d}{dx}(\frac{x^2}{2}) + \frac{d}{dx}(x) + \frac{d}{dx}(1)$

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

$f'(x) = \frac{100 x^{100-1}}{100} + \frac{99 x^{99-1}}{99} + ......... + \frac{2x^{2-1}}{2} + 1.x^{1-1} + \frac{d}{dx}(1)\\ f'(x) = \frac{100 x^{99}}{100} + \frac{99 x^{98}}{99} + ......... + \frac{2x^{1}}{2} + 1.x^{0} + 0\\ f'(x) = x^{99} + x^{98} + ......... +x^{1} + 1 + 0$

Now, then

$f'(1) = 1^{99} + 1^{98} + ......... +1^{1} + 1 + 0 = 100\\ f'(0) = 0^{99} + 0^{98} + ......... +0^{1} + 1 + 0 = 1$

Hence, we conclude that

f'(1) = 100 f'(0)

问题6.求出x ⁿ + ax ^n-1 + a ² x ^n-2 +………….. + a ^n-1 x + a ⁿ对于某个固定实数a的导数。

解决方案：

Given,

f(x) = xⁿ + ax^n-1 + a²x^n-2 + ……………….+ a^n-1x + aⁿ

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

By using this, taking derivative both sides

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(x^n + ax^{n-1} + a^2x^{n-2} + ...................+ a^{n-1}x + a^n)\\ f'(x) = \frac{d}{dx}(x^n) + \frac{d}{dx}(ax^{n-1}) + \frac{d}{dx}(a^2x^{n-2}) + ......... + \frac{d}{dx}(a^{n-1}x) + \frac{d}{dx}(a^n)\\ f'(x) = \frac{d}{dx}(x^n) + a\frac{d}{dx}(x^{n-1}) + a^2\frac{d}{dx}(x^{n-2}) + ......... + a^{n-1}\frac{d}{dx}(x) + a^n\frac{d}{dx}(1)\\ f'(x) = (nx^{n-1}) + a((n-1)x^{n-1-1}) + a^2((n-2)x^{n-2-1}) + ......... + a^{n-1}(1.(x)^{1-1}) + a^n(0)\\ f'(x) = (nx^{n-1}) + a((n-1)x^{n-2}) + a^2((n-2)x^{n-3}) + ......... + a^{n-1}(1) +0\\ f'(x) = nx^{n-1} + a(n-1)x^{n-2} + a^2(n-2)x^{n-3} + ......... + a^{n-1}$

问题7.对于某些常数a和b，找到

(i)(xa)(xb)

解决方案：

f(x) = (x-a) (x-b)

f(x) = x² – (a+b)x + ab

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(x^2 - (a+b)x + ab)\\ f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}((a+b)x) + \frac{d}{dx}(ab)$

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

$f'(x) = (2x^{2-1}) - (a+b)\frac{d}{dx}(x) + ab\frac{d}{dx}(1)\\ f'(x) = 2x^{1} - (a+b)(1x^{1-1}) + ab(0)\\ f'(x) = 2x - (a+b)(x^{0}) + 0\\ f'(x) = 2x - a - b$

(ii)(ax ² + b) ²

解决方案：

f(x) = (ax² + b)²

f(x) = (ax²)² + 2(ax²)(b) + b²

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}((ax^2)^2 + 2(ax^2)(b) + b^2)$

$f'(x) = \frac{d}{dx}((ax^2)^2) + \frac{d}{dx}(2(ax^2)(b)) + \frac{d}{dx}(b^2)\\ f'(x) = \frac{d}{dx}(a^2x^4) + \frac{d}{dx}(2abx^2) + b^2\frac{d}{dx}(1)\\ f'(x) = a^2\frac{d}{dx}(x^4) + 2ab\frac{d}{dx}(x^2) + b^2(0)$

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

$f'(x) = a^2(4x^{4-1}) + 2ab(2x^{2-1}) + 0\\ f'(x) = a^2(4x^3) + 2ab(2x^1) + 0\\ f'(x) = 4a^2x^3 + 2ab(2x) + 0\\ f'(x) = 4a^2x^3 + 4abx$

(iii) $\frac{x-a}{x-b}$

解决方案：

$f(x) = \frac{x-a}{x-b}$

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{x-a}{x-b})$

Using quotient rule, we have

$(\frac{u}{v})' = \frac{uv'-vu'}{u^2}\\ f'(x) = (\frac{(x-b)\frac{d}{dx}(x-a)-(x-a)\frac{d}{dx}(x-b)}{(x-b)^2})\\ f'(x) = (\frac{(x-b)(1)-(x-a)(1)}{(x-b)^2})\\ f'(x) = (\frac{(x-b-x+a)}{(x-b)^2})\\ f'(x) = (\frac{(a-b)}{(x-b)^2})$

问题8.求出$\frac{x^n-a^n}{x-a}$对于一些常数

解决方案：

$f(x) = \frac{x^n-a^n}{x-a}$

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{x^n-a^n}{x-a})$

Using quotient rule, we have

$(\frac{u}{v})' = \frac{uv'-vu'}{u^2}\\ f'(x) = (\frac{(x-a)\frac{d}{dx}(x^n-a^n)-(x^n-a^n)\frac{d}{dx}(x-a)}{(x-a)^2})\\ f'(x) = (\frac{(x-a)[\frac{d}{dx}(x^n)-\frac{d}{dx}(a^n)]-(x^n-a^n)(1)}{(x-a)^2})$

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

$f'(x) = (\frac{(x-a)[(nx^{n-1})-0)]-(x^n-a^n)}{(x-a)^2})\\ f'(x) = (\frac{(x-a)(nx^{n-1})-x^n+a^n}{(x-a)^2})\\ f'(x) = (\frac{(nx^{n-1+1}-anx^{n-1})-x^n+a^n}{(x-a)^2})\\ f'(x) = (\frac{(nx^n-anx^{n-1})-x^n+a^n}{(x-a)^2})$

问题9.求出

(一世) $2x-\frac{3}{4}$

解决方案：

$f(x) = 2x-\frac{3}{4}$

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(2x-\frac{3}{4}) f'(x) = \frac{d}{dx}(2x)-\frac{d}{dx}(\frac{3}{4})$

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

f'(x) = (2x⁰)-0

f'(x) = 2

(ii)(5x ³ + 3x – 1)(x-1)

解决方案：

f(x) = (5x³ + 3x – 1)(x-1)

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}((5x^3 + 3x - 1)(x-1))$

Using product rule, we have

(uv)’ = uv’ + u’v

$f'(x) = (5x^3 + 3x - 1)\frac{d}{dx}(x-1) + (x-1)\frac{d}{dx}(5x^3 + 3x - 1)$

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

$f'(x) = (5x^3 + 3x - 1)(1) + (x-1)((3)5x^{3-1} + 3x^0 - 0)\\ f'(x) = (5x^3 + 3x - 1) + (x-1)(15x^2 + 3)\\ f'(x) = (5x^3 + 3x - 1) + (15x^3 + 3x-(15x^2)-3) \\ f'(x) = (5x^3 + 3x - 1) + (15x^3 + 3x-15x^2-3) \\ f'(x) = 20x^3 - 15x^2 + 6x - 4$

(iii)x ^-3 (5 + 3x)

解决方案：

f(x) = x^-3 (5+3x)

Taking derivative both sides,

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

Using product rule, we have

(uv)’ = uv’ + u’v

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

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

$f'(x) = (x-3)(3) + (5+3x)(-3x^{-3-1})\\ f'(x) = (3x^{-3})+ (5+3x)(-3x^{-4})\\ f'(x) = (3x^{-3})+ (-15x^{-4}+3x(-3x^{-4}))\\ f'(x) = (3x^{-3})- 15x^{-4}-9x^{-4+1})\\ f'(x) = (3x^{-3}) -15x^{-4}-9x^{-3}\\ f'(x) = -6x^{-3} -15x^{-4}$

(iv)x ⁵ (3-6x ^-9 )

解决方案：

f(x) = x⁵ (3-6x^-9)

Taking derivative both sides,

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

Using product rule, we have

(uv)’ = uv’ + u’v

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

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

$f'(x) = (x^5)[\frac{d}{dx}(3)-\frac{d}{dx}(6x^{-9})] + (3-6x^{-9})(5x^{5-1})\\ f'(x) = (x^5)[0-((-9)6x^{-9-1})] + (3-6x^{-9})(5x^{4})\\ f'(x) = (x^5)(54x^{-10}) + (3(5x^{4})-6x^{-9}(5x^{4}))\\ f'(x) = 54x^{-10+5} + (15x^{4} -30x^{-9+4})\\ f'(x) = 54x^{-5} + 15x^{4} -30x^{-5}\\ f'(x) = 24x^{-5} + 15x^{4}$

(v)x ^-4 (3-4x ^-5 )

解决方案：

f(x) = x^-4 (3-4x^-5)

Taking derivative both sides,

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

Using product rule, we have

(uv)’ = uv’ + u’v

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

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

$f'(x) = (x^{-4})[\frac{d}{dx}(3)-\frac{d}{dx}(4x^{-5})] + (3-4x^{-5})(-4x^{-4-1})\\ f'(x) = (x^{-4})[0-(4(-5)x^{-5-1})] + (3-4x^{-5})(-4x^{-5})\\ f'(x) = (x^{-4})[20x^{-6})] + (3(-4x^{-5})-4x^{-5}(-4x^{-5}))\\ f'(x) = (20x^{-6-4}) + (-12x^{-5}-16x^{-5-5})\\ f'(x) = (20x^{-10}) - 12x^{-3} - 16x^{-12})\\ f'(x) = 36x^{-10} - 12x^{-3}$

(六) $\frac{2}{x+1} - \frac{x^2}{3x-1}$

解决方案：

$f(x) = \frac{2}{x+1} - \frac{x^2}{3x-1}$

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{2}{x+1} - \frac{x^2}{3x-1})$

Using quotient rule, we have

$(\frac{u}{v})' = \frac{uv'-vu'}{u^2}\\ f'(x) = [\frac{(x+1)\frac{d}{dx}(2)-(2)\frac{d}{dx}(x+1)}{(x+1)^2})] - [\frac{(3x-1)\frac{d}{dx}(x^2)-(x^2)\frac{d}{dx}(3x-1)}{(3x-1)^2})]$

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

$f'(x) = [\frac{(x+1)(0)-(2)(1)}{(x+1)^2})] - [\frac{(3x-1)(2x^{2-1})-(x^2)(3)}{(3x-1)^2})]\\ f'(x) = [\frac{-2}{(x+1)^2})] - [\frac{(3x-1)(2x)-(x^2)(3)}{(3x-1)^2})]\\ f'(x) = [\frac{-2}{(x+1)^2})] - [\frac{(6x^2-2x)-3x^2)}{(3x-1)^2})]\\ f'(x) = \frac{-2}{(x+1)^2}) - \frac{(3x^2-2x)}{(3x-1)^2})$

问题10：从第一原理求出cos x的导数。

解决方案：

Here, f(x) = cos x

f(x+h) = cos (x+h)

From the first principle,

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} \frac{cos (x+h)-cos x}{h}$

Using the trigonometric identity,

cos a – cos b = -2 sin $(\frac{a+b}{2})$ sin $(\frac{a-b}{2})$

$f'(x) = \lim_{h \to 0} \frac{-2 sin (\frac{x+h+x}{2}) sin (\frac{x+h-x}{2})}{h}\\ f'(x) = \lim_{h \to 0} \frac{-2 sin (\frac{2x+h}{2}) sin (\frac{h}{2})}{h}\\ f'(x) = \lim_{h \to 0} (-2 sin (\frac{2x+h}{2})) \lim_{h \to 0}\frac{sin (\frac{h}{2})}{h}$

Multiplying and diving by 2,

$f'(x) = \lim_{h \to 0} (-2 sin (\frac{2x+h}{2})) \lim_{h \to 0}\frac{sin (\frac{h}{2})}{h} \times \frac{2}{2}\\ f'(x) = \lim_{h \to 0} (-2 sin (\frac{2x+h}{2})) \lim_{h \to 0}\frac{sin (\frac{h}{2})}{\frac{h}{2}} \times \frac{1}{2}\\ f'(x) = (-sin (\frac{2x+0}{2})) \lim_{h \to 0}\frac{sin (\frac{h}{2})}{\frac{h}{2}}$

f'(x) = -sin (x) (1)

f'(x) = -sin x

问题11.查找以下函数的派生：

(i)罪恶x cos x

解决方案：

f(x) = sin x cos x

f(x+h) = sin (x+h) cos (x+h)

From the first principle,

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} \frac{sin (x+h) cos (x+h)-sin x \hspace{0.1cm}cos x}{h}$

Using the trigonometric identity,

sin A cos B = $\frac{1}{2}$(sin (A+B) + sin(A-B))

$f'(x) = \lim_{h \to 0} \frac{\frac{1}{2}(sin (x+h+x+h) + sin(x+h-(x+h)))-(sin (x+x) + sin(x-x)}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{1}{2}(sin (2x+2h) + sin(0))-(sin 2x + sin(0)}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{1}{2}(sin (2x+2h))-(sin 2x)}{h}\\$

Using the trigonometric identity,

sin A – sin B = 2 cos $(\frac{A+B}{2})$ sin $(\frac{A-B}{2})$

$f'(x) = \lim_{h \to 0} \frac{\frac{1}{2}(2 cos (\frac{2x+2h+2x}{2})sin (\frac{2x+2h-2x}{2})}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{1}{2}(2 cos (2x+h)sin (h)}{h}\\ f'(x) = \lim_{h \to 0} \frac{1}{2}(2 cos (2x+h)) \lim_{h \to 0} \frac{sin (h)}{h}\\ f'(x) = cos (2x+0) (1)\\ f'(x) = cos 2x$

(ii)秒x

解决方案：

f(x) = sec x = $\frac{1}{cos x}$

$f(x+h) = \frac{1}{cos (x+h)}$

From the first principle,

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{1}{cos (x+h)}-\frac{1}{cos x}}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{cos x-cos (x+h)}{cos (x+h)cos x}}{h}\\ f'(x) = \frac{1}{cos x}\lim_{h \to 0} \frac{\frac{cos x-cos (x+h)}{cos (x+h)}}{h}$

Using the trigonometric identity,

cos a – cos b = -2 sin $(\frac{a+b}{2})$ sin $(\frac{a-b}{2})$

$f'(x) = \frac{1}{cos x}\lim_{h \to 0} \frac{\frac{-2 sin (\frac{x+x+h}{2}) sin (\frac{x-(x+h)}{2})}{cos (x+h)}}{h}\\ f'(x) = \frac{1}{cos x}\lim_{h \to 0} \frac{-2 sin (\frac{2x+h}{2}) sin (\frac{-h}{2})}{hcos (x+h)}\\ f'(x) = \frac{2}{cos x}\lim_{h \to 0} \frac{sin (\frac{2x+h}{2}) sin (\frac{h}{2})}{hcos (x+h)}\\ f'(x) = \frac{2}{cos x}\lim_{h \to 0} \frac{sin (\frac{2x+h}{2})}{cos (x+h)} \lim_{h \to 0} \frac{sin (\frac{h}{2})}{h}$

Multiply and divide by 2, we have

$f'(x) = \frac{2}{cos x}\lim_{h \to 0} \frac{sin (\frac{2x+h}{2})}{cos (x+h)} \lim_{h \to 0} \frac{sin (\frac{h}{2})}{h} \times \frac{2}{2}\\ f'(x) = \frac{2}{cos x} \frac{sin (\frac{2x+0}{2})}{cos (x+0)} \lim_{h \to 0} \frac{sin (\frac{h}{2})}{\frac{h}{2}} \times \frac{1}{2}\\ f'(x) = \frac{1}{cos x}(\frac{sin (x)}{cos (x)}) \lim_{h \to 0} \frac{sin (\frac{h}{2})}{\frac{h}{2}}\\ f'(x) = \frac{tan x}{cos x}(1) \\ f'(x) = tan x \hspace{0.1cm}sec x$

(iii)5秒x + 4 cos x

解决方案：

f(x) = 5 sec x + 4 cos x

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(5 sec x + 4 cos x)\\ f'(x) = \frac{d}{dx}(5 sec x) + \frac{d}{dx}(4 cos x)\\ f'(x) = 5\frac{d}{dx}(sec x) + 4 \frac{d}{dx}(cos x)$

f'(x) = 5 (tan x sec x) + 4 (-sin x)

f'(x) = 5 tan x sec x – 4 sin x

(iv)cosec x

解决方案：

f(x) = cosec x = $\frac{1}{sin x}$

$f(x+h) = \frac{1}{sin (x+h)}$

From the first principle,

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{1}{sin (x+h)}-\frac{1}{sin x}}{h}\\ f'(x) = \lim_{h \to 0} \frac{\frac{sin x-sin (x+h)}{sin (x+h)sin x}}{h}$

Using the trigonometric identity,

sin a – sin b = 2 cos $(\frac{a+b}{2})$ sin $(\frac{a-b}{2})$

$f'(x) = \frac{1}{sin x}\lim_{h \to 0} \frac{\frac{2 cos (\frac{x+x+h}{2}) sin (\frac{x-(x+h)}{2})}{sin (x+h)}}{h}\\ f'(x) = \frac{1}{sin x}\lim_{h \to 0} \frac{2 cos (\frac{2x+h}{2}) sin (\frac{-h}{2})}{hsin (x+h)}\\ f'(x) = \frac{2}{sin x}\lim_{h \to 0} \frac{cos (\frac{2x+h}{2}) (-sin (\frac{h}{2})}{hsin (x+h)}\\ f'(x) = \frac{-2}{sin x}\lim_{h \to 0} \frac{cos (\frac{2x+h}{2})}{sin (x+h)} \lim_{h \to 0} \frac{sin (\frac{h}{2})}{h}$

Multiply and divide by 2, we have

$f'(x) = \frac{-2}{sin x}\lim_{h \to 0} \frac{cos (\frac{2x+h}{2})}{sin (x+h)}\lim_{h \to 0} \frac{sin (\frac{h}{2})}{h}\times \frac{2}{2}\\ f'(x) = \frac{-2}{sin x} \frac{cos (\frac{2x+0}{2})}{sin (x+0)} \lim_{h \to 0} \frac{sin (\frac{h}{2})}{\frac{h}{2}} \times \frac{1}{2}\\ f'(x) = \frac{-1}{sin x}(\frac{cos (x)}{sin (x)}) \lim_{h \to 0} \frac{sin (\frac{h}{2})}{\frac{h}{2}}\\ f'(x) = \frac{-cot x}{sin x}(1)\\ f'(x) = -cot x\hspace{0.1cm} cosec x$

(v)3张婴儿床x + 5 sec x

解决方案：

f(x) = 3 cot x + 5 cosec x

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(3 cot x + 5 cosec x)$

$f'(x) = \frac{d}{dx}(3 cot\hspace{0.1cm} x) + \frac{d}{dx}(5 cosec\hspace{0.1cm} x)$

f'(x) = 3 g'(x) + 5 $\frac{d}{dx}(cosec \hspace{0.1cm}x)$

Here, 

g(x) = cot x = $\frac{cos x}{sin x}$

$g(x+h) = \frac{cos (x+h)}{sin (x+h)}$

From the first principle,

$g'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ g'(x) = \lim_{h \to 0} \frac{\frac{cos (x+h)}{sin (x+h)}-\frac{cos x}{sin x}}{h}\\ g'(x) = \lim_{h \to 0} \frac{\frac{sin x cos(x+h)-cos x sin (x+h)}{sin (x+h)sin x}}{h}$

Using the trigonometric identity,

sin a cos b – cos a sin b = sin (a-b)

$g'(x) = \lim_{h \to 0} \frac{\frac{sin (x -(x+h))}{sin (x+h)sin x}}{h}\\ g'(x) = \lim_{h \to 0} \frac{sin (-h)}{h sin (x+h)sin x}\\ g'(x) = \lim_{h \to 0} \frac{-sin h}{h sin (x+h)sin x}\\ g'(x) = \frac{-1}{sin x} (\lim_{h \to 0} \frac{1}{sin(x+h)}) (\lim_{h \to 0} \frac{sin h}{h})\\ g'(x) = \frac{-1}{sin x} \frac{1}{sin(x+0)} (1)\\ g'(x) = \frac{-1}{sin^2 x}\\ g'(x) = - cosec^2x$

So, now

f'(x) = 3 g'(x) + 5 $\frac{d}{dx}(cosec\hspace{0.1cm} x)$

f'(x) = 3 (- cosec² x) + 5 (-cot x cosec x)

f'(x) = – 3cosec² x – 5 cot x cosec x

(vi)5罪x – 6 cos x + 7

解决方案：

f(x) = 5 sin x – 6 cos x + 7

f(x+h) = 5 sin (x+h) – 6 cos (x+h) + 7

From the first principle,

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} \frac{5 sin (x+h) - 6 cos (x+h) + 7-(5 sin x - 6 cos x + 7)}{h}\\ f'(x) = \lim_{h \to 0} \frac{5 sin (x+h) - 6 cos (x+h) + 7 - 5 sin x + 6 cos x - 7}{h}\\ f'(x) = \lim_{h \to 0} \frac{5 (sin (x+h) - sin x) - 6 (cos (x+h) - cos x) + 7 - 7}{h}\\ f'(x) = \lim_{h \to 0} \frac{5 (sin (x+h) - sin x) - 6 (cos (x+h) - cos x)}{h}$

Using the trigonometric identity,

sin a – sin b = 2 cos $(\frac{a+b}{2})$ sin $(\frac{a-b}{2})$

cos a – cos b = -2 sin $(\frac{a+b}{2})$ sin $(\frac{a-b}{2})$

$f'(x) = \lim_{h \to 0} \frac{5 (2 cos (\frac{x+h+x}{2}) sin (\frac{x+h-x}{2})) - 6 (-2 sin (\frac{x+h+x}{2}) sin (\frac{x+h-x}{2}))}{h}\\ f'(x) = \lim_{h \to 0} \frac{5 (2 cos (\frac{2x+h}{2}) sin (\frac{h}{2})) - 6 (-2 sin (\frac{2x+h}{2}) sin (\frac{h}{2}))}{h}\\ f'(x) = \lim_{h \to 0} (\frac{10 cos (\frac{2x+h}{2}) sin (\frac{h}{2}) + 12 sin (\frac{2x+h}{2}) sin (\frac{h}{2}))}{h})\\ f'(x) = 10 \lim_{h \to 0} \frac{cos (\frac{2x+h}{2}) sin (\frac{h}{2})}{h} + 12 \lim_{h \to 0} (\frac{sin (\frac{2x+h}{2}) sin (\frac{h}{2}))}{h})$

Multiply and divide by 2, we get

$f'(x) = \frac{2}{2}[10 \lim_{h \to 0} \frac{cos (\frac{2x+h}{2}) sin (\frac{h}{2})}{h} + 12 \lim_{h \to 0} (\frac{sin (\frac{2x+h}{2}) sin (\frac{h}{2}))}{h})]\\ f'(x) = \frac{1}{2}[10 \lim_{h \to 0} \frac{cos (\frac{2x+h}{2}) sin (\frac{h}{2})}{\frac{h}{2}} + 12 \lim_{h \to 0} (\frac{sin (\frac{2x+h}{2}) sin (\frac{h}{2}))}{\frac{h}{2}})]\\ f'(x) = 5 cos (\frac{2x+0}{2}) \lim_{h \to 0}\frac{ sin (\frac{h}{2})}{\frac{h}{2}} + 6 (sin (\frac{2x+0}{2}) \lim_{h \to 0} \frac{sin (\frac{h}{2}))}{\frac{h}{2}})$

f'(x) = 5 cos x (1) + 6  sin x (1)

f'(x) = 5 cos x + 6  sin x 

(vii)2棕褐色x – 7秒x

解决方案：

f(x) = 2 tan x – 7 sec x 

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(2\hspace{0.1cm} tan \hspace{0.1cm}x - 7\hspace{0.1cm} sec\hspace{0.1cm} x )$

f'(x) = $\frac{d}{dx}(2 \hspace{0.1cm}tan \hspace{0.1cm}x) - \frac{d}{dx}(7\hspace{0.1cm} sec\hspace{0.1cm} x)$

f'(x) = 2 g'(x) – 7 $\frac{d}{dx}(sec\hspace{0.1cm} x)$

Here,

g(x) = tan x = $\frac{sin \hspace{0.1cm}x}{cos \hspace{0.1cm}x}$

$g(x+h) = \frac{sin (x+h)}{cos (x+h)}$

From the first principle,

$g'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ g'(x) = \lim_{h \to 0} \frac{\frac{sin (x+h)}{cos (x+h)}-\frac{sin\hspace{0.1cm} x}{cos\hspace{0.1cm} x}}{h}\\ g'(x) = \lim_{h \to 0} \frac{\frac{cos \hspace{0.1cm}x \hspace{0.1cm}sin (x+h)-sin\hspace{0.1cm} x \hspace{0.1cm}cos(x+h)}{cos (x+h)cos \hspace{0.1cm}x}}{h}$

Using the trigonometric identity,

sin a cos b – cos a sin b = sin (a-b)

$g'(x) = \lim_{h \to 0} \frac{\frac{sin (x+h -x)}{cos (x+h)cos x}}{h}\\ g'(x) = \lim_{h \to 0} \frac{sin (h)}{h \hspace{0.1cm}cos (x+h)\hspace{0.1cm}cos\hspace{0.1cm} x}\\ g'(x) = \frac{1}{cos\hspace{0.1cm} x} (\lim_{h \to 0} \frac{1}{cos(x+h)}) (\lim_{h \to 0} \frac{sin h}{h})\\ g'(x) = \frac{1}{cos \hspace{0.1cm}x} \frac{1}{cos(x+0)} (1)\\ g'(x) = \frac{1}{cos^2 x}$

g'(x) = sec²x

So, now

f'(x) = 2 g'(x) – 7 $\frac{d}{dx}(sec \hspace{0.1cm} x)$

f'(x) = 2 (sec²x) – 7 (sec x tan x)

f'(x) = 2sec²x – 7 sec x tan x