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1

问题1：从第一原理中得出以下函数的派生：

(i)-x

解决方案：

f(x) = -x

f(x+h) = -(x+h)

From the first principle,

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

f'(x) = -1

(ii)(-x) ^-1

解决方案：

f(x) = (-x)^-1 = $\frac{-1}{x}$

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

From the first principle,

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

(iii)罪(x + 1)

解决方案：

f(x) = sin(x+1)

f(x+h) = sin((x+h)+1)

From the first principle,

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

Using the trigonometric identity,

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

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

Multiply and divide by 2, we have

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

f'(x) = cos (x+1) (1)

f'(x) = cos (x+1)

(iv) $cos(x-\frac{\pi}{8})$

解决方案：

Here, $f(x) = cos(x-\frac{\pi}{8})\\ f(x+h) = cos((x+h)-\frac{\pi}{8})$

From the first principle,

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} \frac{cos((x+h)-\frac{\pi}{8})-cos(x-\frac{\pi}{8})}{h}$

Using the trigonometric identity,

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

$f'(x) = \lim_{h \to 0} \frac{-2 sin (\frac{(x+h)-\frac{\pi}{8}+x-\frac{\pi}{8}}{2}) sin (\frac{(x+h)-\frac{\pi}{8}-(x-\frac{\pi}{8})}{2})}{h}\\ f'(x) = \lim_{h \to 0} \frac{-2 sin (\frac{(2x+h)-\frac{2\pi}{8}}{2}) sin (\frac{(x+h)-\frac{\pi}{8}-x+\frac{\pi}{8})}{2})}{h}\\ f'(x) = \lim_{h \to 0} \frac{-2 sin (\frac{2x+h-\frac{\pi}{4}}{2}) sin (\frac{h}{2})}{h}$

Multiplying and diving by 2,

$f'(x) = \lim_{h \to 0} \frac{-2 sin (\frac{2x+h-\frac{\pi}{4}}{2}) sin (\frac{h}{2})}{h} \times \frac{2}{2}\\ f'(x) = \lim_{h \to 0} (-2 sin (\frac{2x+h-\frac{\pi}{4}}{2})) \lim_{h \to 0}\frac{sin (\frac{h}{2})}{\frac{h}{2}} \times \frac{1}{2}\\ f'(x) = (-sin (\frac{2x+0-\frac{\pi}{4}}{2})) \lim_{h \to 0}\frac{sin (\frac{h}{2})}{\frac{h}{2}}\\ f'(x) = (-sin (x-\frac{\pi}{8})) (1)\\ f'(x) = -sin (x-\frac{\pi}{8})$

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2：(x + a)

解决方案：

f(x) = x+a

Taking derivative both sides,

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

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

$f'(x) = (1.x^{1-1})+0\\ f'(x) = (1.x^0)+0\\ f'(x) = 1$

问题3： $(px+q) (\frac{r}{s}+s)$

解决方案：

$f(x) = (px+q) (\frac{r}{s}+s)$

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}((px+q) (\frac{r}{s}+s))$

Using the product rule, we have

(uv)’ = uv’+u’v

$f'(x) = (px+q) \frac{d}{dx}(\frac{r}{s}+s) + (\frac{r}{s}+s) \frac{d}{dx}(px+q)$

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

$f'(x) = (px+q) (0) + (\frac{r}{s}+s) (\frac{d}{dx}(px)+\frac{d}{dx}(q))\\ f'(x) = 0 + (\frac{r}{s}+s) (p+0)\\ f'(x) = p(\frac{r}{s}+s)$

问题4：(ax + b)(cx + d) ²

解决方案：

f(x) = (ax+b) (cx+d)²

Taking derivative both sides,

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

Using the product rule, we have

(uv)’ = uv’+u’v

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

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

$f'(x) = (ax+b) \frac{d}{dx}(c^2x^2+2(cx)(d)+d^2) + (cx+d)^2 (\frac{d}{dx}(ax)+\frac{d}{dx}(b))\\ f'(x) = (ax+b) (c^2(2x)+2cd+0) + (cx+d)^2 (a+0)\\ f'(x) = (ax+b) (2xc^2+2cd) + a(cx+d)^2$

问题5： $\frac{ax+b}{cx+d}$

解决方案：

$f(x) = \frac{ax+b}{cx+d}$

Taking derivative both sides,

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

Using the quotient rule, we have

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

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

$f'(x) = \frac{(cx+d) (a) - (ax+b)(c)}{(cx+d)^2}\\ f'(x) = \frac{acx+ad - acx-bc}{(cx+d)^2}\\ f'(x) = \frac{ad-bc}{(cx+d)^2}$

问题6： $\frac{1+\frac{1}{x}}{1-\frac{1}{x}}$

解决方案：

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

Taking derivative both sides,

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

Using the quotient rule, we have

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

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

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

问题7： $\frac{1}{ax^2+bx+c}$

解决方案：

$f(x) = \frac{1}{ax^2+bx+c}$

Taking derivative both sides,

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

Using the quotient rule, we have

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

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

$f'(x) = \frac{(ax^2+bx+c) (0) - (1)(a(2x)+b+0)}{(ax^2+bx+c)^2}\\ f'(x) = \frac{0 - (2ax+b+0)}{(ax^2+bx+c)^2}\\ f'(x) = \frac{- (2ax+b+0)}{(ax^2+bx+c)^2}$

问题8： $\frac{ax+b}{px^2+qx+r}$

解决方案：

$f(x) = \frac{ax+b}{px^2+qx+r}$

Taking derivative both sides,

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

Using the quotient rule, we have

$(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(px^2+qx+r) \frac{d}{dx}(ax+b) - (ax+b)\frac{d}{dx}(px^2+qx+r)}{(px^2+qx+r)^2}$

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

$f'(x) = \frac{(px^2+qx+r) (a+0) - (ax+b)\frac{d}{dx}(p(2x)+q+0)}{(px^2+qx+r)^2}\\ f'(x) = \frac{a(px^2+qx+r) - (ax+b)(2px+q)}{(px^2+qx+r)^2}\\ f'(x) = \frac{apx^2+qax+ra - (2apx^2+qax+2bpx+bq)}{(px^2+qx+r)^2}\\ f'(x) = \frac{apx^2+qax+ra - 2apx^2-qax-2bpx-bq)}{(px^2+qx+r)^2}\\ f'(x) = \frac{ra - apx^2-2bpx-bq)}{(px^2+qx+r)^2}$

问题9： $\frac{px^2+qx+r}{ax+b}$

解决方案：

$f(x) = \frac{px^2+qx+r}{ax+b}$

Taking derivative both sides,

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

Using the quotient rule, we have

$(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(ax+b) \frac{d}{dx}(px^2+qx+r) - (px^2+qx+r)\frac{d}{dx}(ax+b)}{(ax+b)^2}$

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

$f'(x) = \frac{(ax+b) (p(2x)+q+0) - (px^2+qx+r)(a+0)}{(ax+b)^2}\\ f'(x) = \frac{(ax+b) (2px+q) - (px^2+qx+r)(a)}{(ax+b)^2}\\ f'(x) = \frac{2apx^2+qax+2bpx+bq - (apx^2+qax+ra}{(ax+b)^2}\\ f'(x) = \frac{2apx^2+qax+2bpx+bq - apx^2-qax-ra}{(ax+b)^2}\\ f'(x) = \frac{2apx^2+qax+2bpx+bq - apx^2-qax-ra}{(ax+b)^2}\\ f'(x) = \frac{apx^2 - ra +2bpx+bq)}{(ax+b)^2}$

问题10： $\frac{a}{x^4}-\frac{b}{x^2}+cos x$

解决方案：

$f(x) = \frac{a}{x^4}-\frac{b}{x^2}+cos x$

Taking derivative both sides,

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{a}{x^4}-\frac{b}{x^2}+cos x)\\ f'(x) = \frac{d}{dx}(\frac{a}{x^4})-\frac{d}{dx}(\frac{b}{x^2})+\frac{d}{dx}(cos x)$

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

$f'(x) = a \frac{d}{dx}(x^{-4}) - b\frac{d}{dx}(x^{-2})+(- sin x)\\ f'(x) = a (-4x^{-4-1}) - b((-2)x^{-2-1})+(- sin x)\\ f'(x) = a (-4x^{-5}) - b((-2)x^{-3})+(- sin x)\\ f'(x) = -[4ax^{-5} - 2bx^{-3} + sin x]$

问题11： $4\sqrt{x} - 2$

解决方案：

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

Taking derivative both sides,

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

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

$f'(x) = 4 \frac{d}{dx}(x^{\frac{1}{2}})-0\\ f'(x) = 4 (\frac{1}{2}x^{\frac{1}{2}-1})\\ f'(x) = 2 (x^{-\frac{1}{2}})\\ f'(x) = \frac{2}{\sqrt{x}}$

问题12：(ax + b) ⁿ

解决方案：

f(x) = (ax+b)ⁿ

f(x+h) = (a(x+h)+b)ⁿ

f(x+h) = (ax+ah+b)ⁿ

From the first principle,

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-(ax+b)^n}{h}\\ f'(x) = \lim_{h \to 0} \frac{(ax+ah+b)^n-(ax+b)^n}{h}\\ f'(x) = \lim_{h \to 0} \frac{(ax+b)^n (1+\frac{ah}{ax+b})^n-(ax+b)^n}{h}\\ f'(x) = (ax+b)^n \lim_{h \to 0} \frac{(1+\frac{ah}{ax+b})^n-1}{h}$

Using the binomial expansion, we have

$f'(x) = (ax+b)^n \lim_{h \to 0} \frac{[1+n(\frac{ah}{ax+b}+\frac{n(n-1)}{2!}(\frac{ah}{(ax+b)^2}+.....]-1}{h}\\ f'(x) = (ax+b)^n \lim_{h \to 0} \frac{[n(\frac{ah}{ax+b}+\frac{n(n-1)a^2h^2}{2!(ax+b)^2}+.................]}{h}\\ f'(x) = (ax+b)^n \lim_{h \to 0} [n(\frac{a}{ax+b}+\frac{n(n-1)a^2h}{2!(ax+b)^2}+.................]\\ f'(x) = (ax+b)^n [n(\frac{a}{ax+b}+\frac{n(n-1)a^2(0)}{2!(ax+b)^2}+0+0+0......]\\ f'(x) = (ax+b)^n (\frac{na}{ax+b})\\ f'(x) = na (ax+b)^{n-1}$

问题13：(ax + b) ⁿ (cx + d) ^m

解决方案：

f(x) = (ax+b)ⁿ (cx+d)^m

Taking derivative both sides,

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

Using the product rule, we have

(uv)’ = uv’+u’v

$f'(x) = (ax+b)^n \frac{d}{dx}(cx+d)^m + (cx+d)^m \frac{d}{dx}(ax+b)^n$

Let’s take, g(x) = (cx+d)^m

$g'(x) = \frac{d}{dx}(cx+d)^m$

g(x+h) = (c(x+h)+d)^m

g(x+h) = (cx+ch+d)^m

From the first principle,

$g'(x) = \lim_{h \to 0} \frac{g(x+h)-g(x)}{h}\\ g'(x) = \lim_{h \to 0} \frac{(cx+ch+d)^m-(cx+d)^m}{h}\\ g'(x) = \lim_{h \to 0} \frac{(cx+d)^m (1+\frac{ch}{cx+d})^m-(cx+d)^m}{h}\\ g'(x) = (cx+d)^m \lim_{h \to 0} \frac{(1+\frac{ch}{cx+d})^m-1}{h}$

Using the binomial expansion, we have

$g'(x) = (cx+d)^m \lim_{h \to 0} \frac{[1+m(\frac{ch}{cx+d}+\frac{m(m-1)}{2!}(\frac{ch}{(cx+d})^2+.....)]-1}{h}\\ g'(x) = (cx+d)^m \lim_{h \to 0} \frac{[m(\frac{ch}{cx+d}+\frac{m(nm-1)c^2h^2}{2!(cx+d)^2}+.................)]}{h}\\ g'(x) = (cx+d)^m \lim_{h \to 0} [m(\frac{c}{cx+d}+\frac{m(m-1)c^2h}{2!(cx+d)^2}+.................)]\\ g'(x) = (cx+d)^m [m(\frac{c}{cx+d}+\frac{m(m-1)c^2(0)}{2!(cx+d)^2}+0+0+0......)]\\ g'(x) = (cx+d)^m (\frac{mc}{cx+d})\\ g'(x) = mc (cx+d)^{m-1}$

So, as 

$f'(x) = (ax+b)^n \frac{d}{dx}(cx+d)^m + (cx+d)^m \frac{d}{dx}(ax+b)^n\\ f'(x) = (ax+b)^n (mc (cx+d)^{m-1}) + (cx+d)^m (na (ax+b)^{n-1})$

问题14：罪(x + a)

解决方案：

f(x) = sin(x+a)

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

From the first principle,

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

Using the trigonometric identity,

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

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

Multiply and divide by 2, we have

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

问题15：cosec x cot x

解决方案：

f(x) = cosec x cot x

Taking derivative both sides,

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

Using the product rule, we have

(uv)’ = uv’+u’v

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

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

f'(x) = – cot² x cosec x – cosec³ x