11类NCERT解决方案-第13章限制和导数–第13章的其他练习|套装2

找到以下函数的导数(可以理解，a，b，c，d，p，q，r和s是固定的非零常数，m和n是整数)：

问题16： $\frac{cos x}{1+sin x}$

解决方案：

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

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{cos x}{1+sin x})$

Using the quotient rule, we have

$(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(1+sin x) \frac{d}{dx}(cos x) - (cos x)\frac{d}{dx}(1+sin x)}{(1+sin x)^2}\\ f'(x) = \frac{(1+sin x) (-sin x) - (cos x)(0+cos x)}{(1+sin x)^2}\\ f'(x) = \frac{(-sinx - sin^2 x)- (cos^2 x)}{(1+sin x)^2}\\ f'(x) = \frac{-sinx - (sin^2 x + cos^2 x)}{(1+sin x)^2}\\ f'(x) = \frac{-sinx - 1}{(1+sin x)^2}\\ f'(x) = \frac{-(sinx + 1)}{(1+sin x)^2}\\ f'(x) = \frac{-1}{(1+sin x)}$

问题17： $\frac{sin x + cos x}{sin x - cos x}$

解决方案：

$f(x) = \frac{sin x + cos x}{sin x - cos x}$

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{sin x + cos x}{sin x - cos x})$

Using the quotient rule, we have

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

问题18： $\frac{sec x - 1}{sec x + 1}$

解决方案：

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

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{1-cos x}{1+cos x})$

Using the quotient rule, we have

$(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(1+cos x) \frac{d}{dx}(1-cos x) - (1-cos x)\frac{d}{dx}(1+cos x)}{(1+cos x)^2}\\ f'(x) = \frac{(1+cos x) (0-(- sin x)) - (1-cos x)(0+(- sin x))}{(1+cos x)^2}\\ f'(x) = \frac{(1+cos x) (sin x) - (1-cos x)(- sin x)}{(1+cos x)^2}\\ f'(x) = \frac{(sin x + sin x \hspace{0.1cm}cos x) + (sin x - sin x\hspace{0.1cm} cos x)}{(1+cos x)^2}\\ f'(x) = \frac{(sin x + sin x)}{(1+cos x)^2}\\ f'(x) = \frac{2 sin x}{(1+cos x)^2}$

问题19：罪恶^的N×

解决方案：

f(x) = sinⁿ x

When n = 1,

f(x) = sin x

$f'(x) = \frac{d}{dx}(sin x) = cos x$

When n = 2,

f(x) = sin² x = sin x sin x 

$f'(x) = \frac{d}{dx}(sin x\hspace{0.1cm} sin x)$

Using the product rule, we have

(uv)’ = uv’+vu’

$f'(x) = (sin x) \frac{d}{dx}(sin x) + (sin x) \frac{d}{dx}(sin x)$

f'(x) = (sin x) (cos x) + (sin x) (cos x) = 2 sin x cos x

When n = 3,

f(x) = sin³ x = sin² x sin x

$f'(x) = \frac{d}{dx}(sin^2\hspace{0.1cm} x sin x)$

Using the product rule, we have

(uv)’ = uv’+vu’

$f'(x) = (sin^2 x) \frac{d}{dx}(sin x) + (sin x) \frac{d}{dx}(sin^2 x)\\ f'(x) = (sin^2 x) (cos x) + (sin x) (2 sin x\hspace{0.1cm} cos x)\\ f'(x) = (sin^2 x \hspace{0.1cm}cos x) + (2 sin^2 x \hspace{0.1cm}cos x)\\ f'(x) = (sin^2 x\hspace{0.1cm} cos x)[1+2]\\ f'(x) = 3 sin^2 x \hspace{0.1cm}cos x$

Pattern w.r.t n is seen here, as follows

$\frac{d}{dx}(sin^n x) = n sin^{n-1}x\hspace{0.1cm} cos x$

Let’s check this statement.

For P(n) = n sin^n-1x cos x

For P(1),

P(1) = 1 sin^1-1x cos x = cos x. Which is true.

n=k

$P(k) = k sin^{k-1}x \hspace{0.1cm}cos x$

n = k+1

$P(k+1) = \frac{d}{dx}(sin^{k+1} x) = \frac{d}{dx}(sin^k x \hspace{0.1cm}sin x)$

Using the product rule, we have

(uv)’ = uv’+vu’

= (sin^k x) $\frac{d}{dx}(sin x)$ + (sin x) $\frac{d}{dx}(sin^k x)$

= (sin^k x) (cos x) + (sin x) (k sin^k-1 x cos x)

= (sin^k x) (cos x)[k+1]

Hence proved for P(k+1).

So,$\frac{d}{dx}(sin^n x) = n sin^{n-1}x \hspace{0.1cm}cos x$ is true.

问题20： $\frac{a+bsin x}{c+dcos x}$

解决方案：

$f(x) = \frac{a+bsin x}{c+dcos x}$

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{a+bsin x}{c+dcos x})$

Using the quotient rule, we have

$(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(c+dcos x) \frac{d}{dx}(a+bsin x) - (a+bsin x)\frac{d}{dx}(c+dcos x)}{(c+dcos x)^2}\\ f'(x) = \frac{(c+dcos x) [\frac{d}{dx}(a)+\frac{d}{dx}(bsin x)] - (a+bsin x)\frac{d}{dx}(c)+\frac{d}{dx}(dcos x)}{(c+dcos x)^2}$

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

$f'(x) = \frac{(c+dcos x) [0+b cos x] - (a+bsin x)[0+(-d sin x)]}{(c+dcos x)^2}\\ f'(x) = \frac{(c+dcos x) (b cos x) + (a+bsin x)(d sin x)}{(c+dcos x)^2}\\ f'(x) = \frac{(bc\hspace{0.1cm} cos x+db\hspace{0.1cm} cos^2 x) + (ad\hspace{0.1cm} sin x+db\hspace{0.1cm} sin^2 x)}{(c+d\hspace{0.1cm}cos x)^2}\\ f'(x) = \frac{(bc\hspace{0.1cm} cos x + db (cos^2 x + sin ^2 x) + ad \hspace{0.1cm}sin x)}{(c+d\hspace{0.1cm}cos x)^2}\\ f'(x) = \frac{(bc\hspace{0.1cm} cos x + db+ ad\hspace{0.1cm} sin x)}{(c+d\hspace{0.1cm}cos x)^2}$

问题21： $\frac{sin(x+a)}{cos x}$

解决方案：

$f(x) = \frac{sin(x+a)}{cos x}$

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{sin(x+a)}{cos x})$

Using the quotient rule, we have

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

Let’s take g(x) = sin (x+a)

$g'(x) = \frac{d}{dx}(sin(x+a))$

g(x+h) = sin((x+h)+a)

From the first principle,

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

Using the trigonometric identity,

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

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

Multiply and divide by 2, we have

$g'(x) = 2 cos (\frac{2x+0+2a}{2})) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{h})) \times \frac{2}{2}\\ g'(x) = 2 cos (x+a) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{\frac{h}{2}})) \times \frac{1}{2}\\ g'(x) = cos (x+a) (1)\\ g'(x) = cos (x+a)$

Hence, 

$f'(x) = \frac{(cos x) \frac{d}{dx}(sin(x+a)) + (sin(x+a))(sin x)}{(cos x)^2}\\ f'(x) = \frac{(cos x) (cos(x+a)) + (sin(x+a))(sin x)}{(cos x)^2}$

Using the trigonometric identity,

cos A cos B + sin A sin B = cos (A-B)

$f'(x) = \frac{cos(x+a-x)}{(cos x)^2}\\ f'(x) = \frac{cos(a)}{(cos x)^2}$

问题22：x ⁴ (5sin x – 3cos x)

解决方案：

f(x) = x⁴ (5sin x – 3cos x)

Taking derivative both sides,

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

Using the product rule, we have

(uv)’ = uv’ + vu’

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

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

$f'(x) = (x^4) [\frac{d}{dx}(5sin x) - \frac{d}{dx}(3cos x)] + (5sin x - 3cos x)(4x^{4-1})$

f'(x) = (x⁴) [(5 cos x) – (3 (- sin x))] + (5sin x – 3cos x)(4x³)

f'(x) = (x⁴) [(5 cos x) + (3 sin x)] + (5sin x – 3cos x)(4x³)

f'(x) = (x³) [5x cos x + 3x sin x + 20sin x – 12 cos x]

问题23：(x ² +1)cos x

解决方案：

f(x) = (x²+1) cos x

Taking derivative both sides,

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

Using the product rule, we have

(uv)’ = uv’ + vu’

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

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

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

f'(x) = (x²+1) (- sin x) + (cos x)[(2x^2-1)+0]

f'(x) = -x² sin x- sin x + 2x cos x

问题24： $(ax^2+sin x) (p+q cos x)$

解决方案：

$f(x) = (ax^2+sin x) (p+q cos x)$

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}((ax^2+sin x) (p+q cos x))$

Using the product rule, we have

(uv)’ = uv’ + vu’

$f'(x) = (ax^2+sin x) \frac{d}{dx}(p+q cos x) + (p+q cos x)\frac{d}{dx}((ax^2+sin x))\\ f'(x) = (ax^2+sin x) [\frac{d}{dx}(p)+\frac{d}{dx}(q cos x)] + (p+q cos x)[\frac{d}{dx}(ax^2)+\frac{d}{dx}(sin x)]$

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

$f'(x) = (ax^2+sin x) [0+q(- sin x)] + (p+q cos x)[a(2x^{2-1})+(cos x)]\\ f'(x) = (ax^2+sin x) (- q sin x) + (p+q cos x)[2ax+cos x]\\ f'(x) = - q sin x(ax^2 + sin x) + (p+q cos x)[2ax+cos x]$

问题25：(x + cos x)(x –棕褐色x)

解决方案：

f(x) = (x + cos x)(x – tan x)

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}((x + cos x)(x - tan x))$

Using the product rule, we have

(uv)’ = uv’ + vu’

$f'(x) = (x + cos x) \frac{d}{dx}(x - tan x) + (x - tan x)\frac{d}{dx}(x + cos x)\\ f'(x) = (x + cos x) [\frac{d}{dx}(x) - \frac{d}{dx}(tan x)] + (x - tan x)[\frac{d}{dx}(x) + \frac{d}{dx}(cos x)]$

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

$f'(x) = (x + cos x) [1 - \frac{d}{dx}(tan x)] + (x - tan x)[1 + (- sin x)]$

Let’s take g(x) = tan x

$g'(x) = \frac{d}{dx}(tan x)\\ g(x) = tan x = \frac{sin \hspace{0.1cm}x}{cos \hspace{0.1cm}x}\\ g(x+h) = \frac{sin (x+h)}{cos (x+h)}$

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

Hence, 

f'(x) = (x + cos x) $[1 - \frac{d}{dx}(tan x)]$ + (x – tan x)[1 + (- sin x)]

f'(x) = (x + cos x) [1 – (sec² x)] + (x – tan x)[1 – sin x]

f'(x) = (x + cos x) [tan² x] + (x – tan x)[1 – sin x]

f'(x) = tan² x(x + cos x) + (x – tan x)[1 – sin x]

问题26： $\frac{4x+5sin x}{3x+7cos x}$

解决方案：

$f(x) = \frac{4x+5sin x}{3x+7cos x}$

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{4x+5sin x}{3x+7cos x})$

Using the quotient rule, we have

$(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(3x+7cos x) \frac{d}{dx}(4x+5sin x) - (4x+5sin x)\frac{d}{dx}(3x+7cos x)}{(3x+7cos x)^2}\\ f'(x) = \frac{(3x+7cos x) [\frac{d}{dx}(4x)+\frac{d}{dx}(5sin x)] - (4x+5sin x)[\frac{d}{dx}(3x)+\frac{d}{dx}(7cos x)]}{(3x+7cos x)^2}$

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

$f'(x) = \frac{(3x+7cos x) [4+(5 cos x)] - (4x+5sin x)[3 + 7(- sin x)]}{(3x+7cos x)^2}\\ f'(x) = \frac{(3x+7cos x) [4+5 cos x] - (4x+5sin x)[3 - 7 sin x]}{(3x+7cos x)^2}\\ f'(x) = \frac{(12x+28 cos x+15x cos x + 35 cos^2x) - [(12x + 15sin x)-(28x sin x + 35 sin^2 x)]}{(3x+7cos x)^2}\\ f'(x) = \frac{(12x+28 cos x+15x cos x + 35 cos^2x)- (12x + 15sin x-28x sin x - 35 sin^2 x)}{(3x+7cos x)^2}\\ f'(x) = \frac{(12x+28 cos x+15x cos x + 35 cos^2x - 12x - 15sin x+28x sin x + 35 sin^2 x)}{(3x+7cos x)^2}\\ f'(x) = \frac{(28 cos x+15x cos x + 35 (cos^2x + sin^2x) - 15sin x + 28x sin x)}{(3x+7cos x)^2}\\ f'(x) = \frac{(28 cos x+15x cos x + 35 - 15sin x + 28x sin x)}{(3x+7cos x)^2}$

问题27： $\frac{x^2 cos(\frac{\pi}{4})}{sin x}$

解决方案：

$f(x) = \frac{x^2 cos(\frac{\pi}{4})}{sin x}$

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{x^2 cos(\frac{\pi}{4})}{sin x})\\ \frac{d}{dx}(f(x)) = cos(\frac{\pi}{4}) \frac{d}{dx}(\frac{x^2)}{sin x})$

Using the quotient rule, we have

$(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = (cos(\frac{\pi}{4}) [\frac{(sin x) \frac{d}{dx}(x^2) - (x^2)\frac{d}{dx}(sin x)}{(sin x)^2}]$

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

$f'(x) = cos(\frac{\pi}{4}) [\frac{(sin x) (2x^{2-1}) - (x^2) (cos x)}{(sin x)^2}]\\ f'(x) = cos(\frac{\pi}{4}) [\frac{2x sin x - x^2 cos x)}{(sin x)^2}]\\ f'(x) = [\frac{(x cos(\frac{\pi}{4})(2 sin x - x cos x)}{(sin x)^2}]$

问题28： $\frac{x}{1+tan x}$

解决方案：

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

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{x}{1+tan x)})$

Using the quotient rule, we have

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

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

$f'(x) = \frac{(1+tan x) (1) - (x)[\frac{d}{dx}(1)+\frac{d}{dx}(tan x)]}{(1+tan x)^2}\\ f'(x) = \frac{(1+tan x) - (x)[0+\frac{d}{dx}(tan x)]}{(1+tan x)^2}$

Let’s take g(x) = tan x

$g'(x) = \frac{d}{dx}(tan x)\\ g(x) = tan x = \frac{sin \hspace{0.1cm}x}{cos \hspace{0.1cm}x}\\ g(x+h) = \frac{sin (x+h)}{cos (x+h)}$

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

Hence, $f'(x) = \frac{(1+tan x) - (x)[0+\frac{d}{dx}(tan x)]}{(1+tan x)^2}\\ f'(x) = \frac{1+tan x - x sec^2x}{(1+tan x)^2}$

问题29：(x +秒x)(x棕褐色x)

解决方案：

f(x) = (x + sec x) (x-tan x)

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}((x + sec x) (x-tan x))$

Using the product rule, we have

(uv)’ = uv’ + vu’

$f'(x) = (x + sec x) \frac{d}{dx}(x - tan x) + (x - tan x)\frac{d}{dx}(x + sec x)\\ f'(x) = (x + sec x) [\frac{d}{dx}(x) - \frac{d}{dx}(tan x)] + (x - tan x)[\frac{d}{dx}(x) + \frac{d}{dx}(sec x)]$

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

$f'(x) = (x + sec x) [1 - \frac{d}{dx}(tan x)] + (x - tan x)[1 + \frac{d}{dx}(sec x)]$

Let’s take g(x) = tan x

$g'(x) = \frac{d}{dx}(tan x)\\ g(x) = tan x = \frac{sin \hspace{0.1cm}x}{cos \hspace{0.1cm}x}\\ g(x+h) = \frac{sin (x+h)}{cos (x+h)}$

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

Now, let’s take h(x) = sec x = $\frac{1}{cos x}$

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

From the first principle,

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

Using the trigonometric identity,

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

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

Multiply and divide by 2, we have

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

Hence, $f'(x) = (x + sec x) [1 - \frac{d}{dx}(tan x)] + (x - tan x)[1 + \frac{d}{dx}(sec x)]\\ f'(x) = (x + sec x) [1 - (sec^2x)] + (x - tan x)[1 + (sec x \hspace{0.1cm}tan x)]\\ f'(x) = (x + sec x) [tan^2x)] + (x - tan x)[1 + (sec x \hspace{0.1cm}tan x)]$

问题30： $\frac{x}{sin^nx}$

解决方案：

$f(x) = \frac{x}{sin^nx}$

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{x}{sin^nx})$

Using the quotient rule, we have

$(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(sin^nx) \frac{d}{dx}(x) - (x)\frac{d}{dx}(sin^nx)}{(sin^nx)^2}$

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

$f'(x) = \frac{(sin^nx) (1) - (x)\frac{d}{dx}(sin^nx)}{(sin^nx)^2}$

Let’s take, g(x) = sinn x

When n = 1,

g(x) = sin x

$g'(x) = \frac{d}{dx}(sin x) = cos x$

When n = 2,

$g(x) = sin^2 x = sin x sin x\\ g'(x) = \frac{d}{dx}(sin x\hspace{0.1cm} sin x)$

Using the product rule, we have

(uv)’ = uv’+vu’

$g'(x) = (sin x) \frac{d}{dx}(sin x) + (sin x) \frac{d}{dx}(sin x)$

g'(x) = (sin x) (cos x) + (sin x) (cos x) = 2 sin x cos x

When n = 3,

g(x) = sin³ x = sin² x sin x

$g'(x) = \frac{d}{dx}(sin^2 x\hspace{0.1cm} sin x)$

Using the product rule, we have

(uv)’ = uv’+vu’

$g'(x) = (sin^2 x) \frac{d}{dx}(sin x) + (sin x) \frac{d}{dx}(sin^2 x)\\ g'(x) = (sin^2 x) (cos x) + (sin x) (2 sin x \hspace{0.1cm}cos x)\\ g'(x) = (sin^2 x \hspace{0.1cm}cos x) + (2 sin^2 x\hspace{0.1cm} cos x)\\ g'(x) = (sin^2 x\hspace{0.1cm} cos x)[1+2]\\ g'(x) = 3 sin^2 x\hspace{0.1cm} cos x$

Pattern w.r.t n is seen here, as follows

$\frac{d}{dx}(sin^n x) = n sin^{n-1}x\hspace{0.1cm} cos x$

Let’s check this statement.

For $P(n) = n sin^{n-1}x \hspace{0.1cm}cos x$

For P(1),

$P(1) = 1 sin^{1-1}x \hspace{0.1cm}cos x = cos x$. Which is true.

n=k

$P(k) = k sin^{k-1}x \hspace{0.1cm}cos x$

n = k+1

$P(k+1) = \frac{d}{dx}(sin^{k+1} x) = \frac{d}{dx}(sin^k x \hspace{0.1cm}sin x)$

Using the product rule, we have

(uv)’ = uv’+vu’

$= (sin^k x) \frac{d}{dx}(sin x) + (sin x) \frac{d}{dx}(sin^k x)\\ = (sin^k x) (cos x) + (sin x) (k sin^{k-1}x \hspace{0.1cm}cos x)\\ = (sin^k x) (cos x)[k+1]$

Hence proved for P(k+1).

So, $\frac{d}{dx}(sin^n x) = n sin^{n-1}x \hspace{0.1cm}cos x$ is true.

So, the given equation will be

$f'(x) = \frac{(sin^nx) (1) - (x)\frac{d}{dx}(sin^nx)}{sin^{2n}x}\\ f'(x) = \frac{(sin^nx) - x(n sin^{n-1}x cos x)}{sin^{2n}x}$