第12类NCERT解决方案-数学第I部分-第5章连续性和可微性-第5章的其他练习

将练习1到11中的函数分开。

问题1.(3 x ² – 9x – 5) ⁹

解决方案：

Let us assume y = (3x² – 9x – 5)⁹

Now, differentiate w.r.t x

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

Using chain rule, we get

= 9(3x² – 9 x + 5)⁸ $\frac{d}{dx}(3x^2 -9x +5)$

= 9(3x² – 9x + 5)⁸.(6x – 9)

= 9(3x ² – 9x + 5)⁸.3(2x – 3)

= 27(3x² – 9x + 5)⁸ (2x – 3)

问题2.罪³ x + cos ⁶ x

解决方案：

Let us assume y = sin³ x + cos⁶ x

Now, differentiate w.r.t x

$\frac{dy}{dx}= \frac{d}{dx}(sin^3x + cos^6x)$

Using chain rule, we get

= $\frac{d}{dx}sin^3x +\frac{d}{dx} cos^6x$

= $3sin^2x.\frac{d}{dx}(sinx)+6sin^2x.\frac{d}{dx}(cosx)$

= 3 sin² x. cos x + 6 cos⁵ x.(-sin x)

= 3 sin x cos x(sin x – 2 cos⁴ x)

问题3. 5x ^{3 cos 2 x}

解决方案：

Let us assume y = 5x^{3 cos 2x}

Now we’re taking logarithm on both the sides

logy = 3 cos 2 x log 5 x

Now, differentiate w.r.t x

$\frac{1}{y}\frac{dy}{dx}=3[log5x.\frac{d}{dx}(cos2x)+cos2x.\frac{d}{dx}(log5x)]$

$\frac{dy}{dx}=3y[log 5x(-sin2x).\frac{d}{dx}(2x)+cos2x.\frac{1}{5x}.\frac{d}{dx}(5x)]$

$\frac{dy}{dx}=3y[-2sin2xlog5x+\frac{cos2x}{x}]$

$\frac{dy}{dx}=3y[\frac{3cos2x}{x}-6sin2xlog5x]$

$\frac{dy}{dx}=5x^{3cos2x}[\frac{3cos2x}{x}-6sin2xlog5x]$

问题4. sin ^-1 (x√x)，0≤x≤1

解决方案：

Let us assume y = sin^-1(x√x)

Now, differentiate w.r.t x

$\frac{dy}{dx}=\frac{d}{dx}sin^{-1}(x\sqrt{x})$

Using chain rule, we get

=$\frac{1}{\sqrt{1-(x\sqrt{x})^2}}.\frac{d}{dx}(x\sqrt{x})$

=$\frac{1}{\sqrt{1-x^3}}.\frac{d}{dx}[x^{\frac{3}{2}}]$

=$\frac{1}{\sqrt{1-x^3}}.\frac{3}{2}.x^{\frac{1}{2}}$

=$\frac{3\sqrt{x}}{2\sqrt{1-x^3}}$

=$\frac{3}{2}\sqrt{\frac{x}{1-x^3}}$

问题5 $\frac{cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}$ ，-2

解决方案：

Let us assume y =$\frac{cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}$

Now, differentiate w.r.t x and by quotient rule, we obtain

$\frac{dy}{dx}=\frac{\sqrt{2x+7}\frac{d}{dx}(cos^{-1}\frac{x}{2})-(cos^{-1}\frac{x}{2})\frac{d}{dx}(\sqrt{2x+7})}{(\sqrt{2x+7})^2}$

= $\frac{\sqrt{2x+7}[\frac{-1}{\sqrt{1-(\frac{x}{2})^2}}.\frac{d}{dx}(\frac{x}{2})]-(cos^{-1}\frac{x}{2})\frac{1}{2\sqrt{2x+7}}.\frac{d}{dx}(2x+7)}{2x+\frac{7}{2}}$

= $\frac{\sqrt{2x+7}\frac{-1}{\sqrt{4-x^2}}-[cos^{-1}\frac{x}{2}]\frac{2}{2\sqrt{2x+7}}}{2x+7}$

= $\frac{-\sqrt{2x+7}}{\sqrt{4-x^2}(2x+7)}-\frac{cos^{-1}\frac{x}{2}}{(\sqrt{2x+7})(2x+7)}$

= $-[\frac{1}{\sqrt{4-x^2}\sqrt{2x+7}}+\frac{cos^{-1}\frac{x}{2}}{(2x+7)^\frac{1}{2}}]$

问题6。 $cot^{-1}\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}$ ，0 π/ 2

解决方案：

Let us assume y = $cot^{-1}\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}$           ……(1)

Now solve $\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}$

= $\frac{(\sqrt{1+sinx}+\sqrt{1-sinx})^2}{(\sqrt{1+sinx}-\sqrt{1-sinx})(\sqrt{1+sinx}+\sqrt{1-sinx})}$

= $\frac{(1+sinx)+(1-sinx)+2\sqrt{(1-sinx)(1+sinx)}}{(1+sinx)-(1-sinx)}$

= $\frac{2+2\sqrt{1-sin^2x}}{2sinx}$

= $\frac{1+cosx}{sinx}$

= $\frac{2cos^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}$

= cotx/2

Now put this value in eq(1), we get

y = cot^-1(cotx/2)

y = x/2

Now, differentiate w.r.t x

$\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}(x)$

dy/dx = 1/2

问题7.(log x) ^{log x} ，x> 1

解决方案：

Let us assume y = (log x)^{log x}

Now we are taking logarithm on both sides,

log y = log x .log(log x)

Now, differentiate w.r.t x on botj side, we get

$\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[logx.log(logx)]$

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

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

$\frac{dy}{dx}=y[\frac{1}{x}log(logx)+\frac{1}{x}]$

$\frac{dy}{dx}=(logx)^{logx}[\frac{1}{x}+\frac{log(logx)}{x}]$

问题8.对于某些常数a和b，cos(a cos x + b sin x)。

解决方案：

Let us assume y = cos(a cos x + b sin x)

Now, differentiate w.r.t x

$\frac{dy}{dx}=\frac{d}{dx}cos(acosx+bsinx)$

By using chain rule, we get

$\frac{dy}{dx}=-sin(acosx+bsinx).\frac{d}{dx}(acosx+bsinx)$

= -sin x(a cos x + b sin x).[a (-sin x) + b cos x]

= (a sin x – b cos x).sin (a cos x + b sin x)

问题9.(sin x – cos x) ^{(sin x – cos x)} ， π/ 4

解决方案：

Let us assume y = (sin x – cos x)^{(sin x – cos x)}

Now we are taking logarithm on both sides,

log y = (sin x – cos x).log(sin x – cos x)

Now, differentiate w.r.t x, we get

$\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[(sinx-cosx)log(sinx-cosx)]$

Using chain rule, we get

$\frac{1}{y}\frac{dy}{dx}=log(sinx-cosx).\frac{d}{dx}(sinx-cosx)+(sinx-cosx).\frac{d}{dx}log(sinx-cosx)$

$\frac{1}{y}\frac{dy}{dx}=log(sinx-cosx).(cosx+sinx)+(sinx-cosx).\frac{1}{(sinx-cosx)}.\frac{d}{dx}(sinx-cosx)$

dy/dx = (sinx – cosx)^{(sinx – cosx)}[(cosx + sinx).log(sinx – cosx) + (cosx + sinx)]

dy/dx = (sinx – cosx)^{(sinx – cosx)}(cosx + sinx)[1 + log (sinx – cosx)]

问题10. x ^x + x ^a + a ^x + ^a对于某些固定的a> 0和x> 0

解决方案：

Let us assume y = x^x + x^a + a^x + a^a

Also, let us assume x^x = u, x^a = v, a^x = w, a^a = s

Therefore, y = u + v + w + s

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

$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}+\frac{ds}{dx}$  ……….(1)

So first we solve: u = x^x

Now we are taking logarithm on both sides,

log u = log x^x

log u = x log x

On differentiating both sides w.r.t x, we get

$\frac{1}{u}\frac{du}{dx}=logx.\frac{d}{dx}(x)+x.\frac{d}{dx}(logx)$

$\frac{du}{dx}=u[logx.1+x.\frac{1}{x}]$

du/dx = x^x[logx + 1] = x^x(1 + logx)    …….(2)

Now we solve: v = x^a

On differentiating both sides w.r.t x, we get

$\frac{dv}{dx}=\frac{d}{dx}(x^a)$

dv/dx = ax^{(a – 1)}   ……(3)

Now we solve: w = a^x

Now we are taking logarithm on both sides,

log w =log a ^x

log w = x log a

On differentiating both sides w.r.t x, we get

$\frac{1}{w}.\frac{dw}{dx}=loga.\frac{d}{dx}(x)$

dw/dx = w loga

dw/dx = a^xloga ………(4)

Now we solve: s = a ^a

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

ds/dx = 0 ………(5)

Now put all these values from eq(2), (3), (4), (5) in eq(1), we get 

dy/dx = x^x(1 + logx) + ax^{(a – 1)} + a^xloga + 0

= x^x (1 + log x) + ax^{a -1} + a^x log a

问题11.区分wrt x， $x^{x^2-3}+(x-3)^{x^2}$ ，对于x> 3

解决方案：

Let us assume y =$x^{x^2-3}+(x-3)^{x^2}$

Also let us considered u = $x^{x^2-3}$and v = $(v-3)^{x^2}$

so, y = u + v

On differentiating both side w.r.t x, we get

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

So, now we solve, u = $x^{x^2-3}$

Now we are taking logarithm on both sides,

log u = log $(x^{x^2-3})$

log u = (x ² – 3) log x

On differentiating w.r.t x, we get 

$\frac{1}{u}.\frac{du}{dx} = logx.\frac{d}{dx}(x^2-3) + (x^2-3).\frac{d}{dx}(log x)$

$\frac{1}{u}.\frac{dy}{dx}= logx.2x+(x^2-3).\frac{1}{x}$

=$\frac{du}{dx}= x^{x^2-3}.[\frac{x^2-3}{x}+ 2xlogx]$      …….(2)

Now we solve: v = $(x-3)^{x^2}$

Now we are taking logarithm on both sides,

log v = $log(x-3)^{x^2}$

log v = x² log(x – 3)

On differentiating both sides w.r.t x, we get 

$\frac{1}{v}.\frac{dv}{dx}= log (x-3).\frac{d}{dx}(x^2)+x^2\frac{d}{dx}[log(x-3)]$

$\frac{1}{v}.\frac{dv}{dx}=log (x-3).2x+x^2.\frac{1}{x-3}.\frac{d}{dx}(x-3)$

$\frac{dv}{dx}=v[2xlog(x-3)+\frac{x^2}{x-3}.1]$

$\frac{dv}{dx}=(x-3)^{x^2}[\frac{x^2}{x-3}+2xlog(x-3)]$      …..(3)

Now put all these values from eq(2), and (3) in eq(1), we get 

$\frac{dy}{dx}= x^{x^2-3}[\frac{x^2-3}{x}+2xlogx]+(x-3)^{x^2}[\frac{x^2}{x-3}+2xlog(x-3)]$

问题12.如果y = 12(1-cos t)，x = 10(t-sin t)，- π/ 2

解决方案：

According to the question

y = 12(1 – cos t)   ……(1)

x = 10 (t – sin t)  ……(2)

So, \frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}    ……(3)

On differentiating eq(1) w.r.t t, we get 

$\frac{dy}{dt} = \frac{d}{dt} [12 (1 - cost)]$ 

= $12\frac{d}{dt}(1- cost )$ 

= 12.[0 – (- sin t)] 

= 12 sin t

On differentiating eq(2) w.r.t t, we get 

$\frac{dx}{dt}=\frac{d}{dt}[10. (t - sin t)]$ 

= $10\frac{d}{dt} (t - sin t)$

= 10(1 – cos t)

Now put the value of dy/dt and dx/dt in eq(3), we get

$\frac{dy}{dx}=\frac{12 sin t}{10(1-cost)}$

= $\frac{12.2sin\frac{t}{2}cos \frac{t}{2}}{10.2sin^2\frac{t}{2}}$      

= 6/5 cot t/2                                      

问题13查找DY / DX，如果y = ^-1罪X +罪^{-1√1-X 2，0}

解决方案：

According to the question

y = sin^-1 x + sin^-1√1 – x²

On differentiating w.r.t x, we get 

$\frac{dy}{dx}=\frac{d}{dx}[sin^{-1}x+sin ^{-1}\sqrt{1-x^2}]$

Using chain rule, we get

$\frac{dy}{dx}=\frac{d}{dx}(sin^{-1}x)+\frac{d}{dx}(sin ^{-1}\sqrt{1-x^2})$

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

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

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

= $\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}$

dy/dx = 0

问题14.如果x√1+ y +y√1+ x = 0，则-1

解决方案：

According to the question

x√1 + y = -y√1 + x 

On squaring both sides, we get

x² (1 + y) = y² (1 + x)

⇒ x² + x² y = y² + x y²

⇒ x² – y² = xy ² – x² y

⇒ x² – y² = xy (y – x)

⇒ (x + y)(x – y) = xy (y – x)

⇒ x + y = -xy

⇒ (1 + x) y = -x

⇒ y = -x/(1 + x)

On differentiating both sides w.r.t x, we get

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

= $\frac{(1+x)-x}{(1+x)^2}$

= $-\frac{1}{(1+x)^2}$

Hence proved.

问题15。如果(x – a) ² +(y – b) ² = c ² ，则对于某些c> 0，证明$\frac{[1+(\frac{dy}{dx})^2]^\frac{3}{2}} {\frac{d^2y}{dx^2}}$是一个独立于a和b的常数。

解决方案：

According to the question

(x – a)²+ (y – b)²= c²

On differentiating both side w.r.t x, we get 

$\frac{d}{dx}[(x-a)^2]+\frac{d}{dx}[(y-b)^2]=\frac{d}{dx}(c^2)$

⇒ 2(x – a).$\frac{d}{dx}(x-a)$ + 2(y – b)$\frac{d}{dx}(y-b)$ = 0 

⇒ 2(x – a).1 + 2(y – b).$\frac{dy}{dx}$= 0

⇒$\frac{dy}{dx}=\frac{-(x-a)}{y-b}$  …….(1)

Again on differentiating both side w.r.t x, we get 

$\frac{d^2y}{dx^2} = \frac{d}{dx}[\frac{-(x-a)}{y-b}]$

$= -[\frac{(y-b).\frac{d}{dx}(x-a)-(x-a).\frac{d}{dx}(y-b)}{(y-b)^2}]$

$=-[\frac{(y-b)-(x-a).\frac{dy}{dx}}{(y-b)^2}]$

$= -[\frac{(y-b)-(x-a).\{-\frac{(x-a)}{y-b}\}}{(y-b)^2}]$ …….[From equation (1)]

$=-[\frac{(y-b)^2+(x-a)^2}{(y-b)^3}]$

$[\frac{1+[\frac{dy}{dx}]^2}{\frac{d^2y}{dx^2}}]^{\frac{3}{2}}= \frac{[1+\frac{(x-a)^2}{(y-b)^2}]^{\frac{3}{2}}}{-[\frac{(y-b)^2 + (x-a)^2}{(y-b)^3}]}$

=$\frac{[\frac{[(y-b)^2+(x-a)^2]}{(y-b)^2}]^{\frac{3}{2}}}{-[\frac{(y-b)^2+(x-a)^2}{(y-b)^3}]}$

=$\frac{[\frac{c^2}{(y-b)^2}]\frac{3}{2}}{-\frac{c^2}{(y-b)^3}}$

=$\frac{\frac{c^3}{(y-b)^3}}{-\frac{c^2}{(y-b)^3}}$

= – c, which is constant and is independent of a and b.

Hence proved.

问题16。如果cos y = x cos(a + y)，且cos a≠±1，则证明$\frac{dy}{dx}= \frac{cos^2(a+y)}{sina}$

解决方案：

According to the question

cos y = x cos (a + y)

On differentiating both side w.r.t x, we get 

$\frac{d}{dx}[cos y]$= $\frac{d}{dx}[x cos(a + y)]$ 

⇒ – sin y dy/dx = cos (a + y). $\frac{d}{dx}(x)$ + x$\frac{d}{dx}[cos (x+y)]$ 

⇒ – sin y dy/dx = cos (a + y) + x [-sin (a + y)]dy/dx

⇒ [x sin (a + y) – sin y] dy/dx = cos (a + y) ……..(1)

Since cos y = x cos (a + y), x =$\frac{cosy }{cos (a+y)}$

Now we can reduce eq(1)

$[\frac{cosy}{cos(a+y)}.sin(a+y)-siny]\frac{dy}{dx}$= cos(a + y)

⇒ [cos y.sin (a + y)- sin y.cos (a + y)].dy/dx = cos²(a + y)

⇒ sin(a + y – y)dy/dx = cos²(a + b)

⇒$\frac{dy}{dx}= \frac{cos^2(a+b)}{sin a}$

Hence proved.

问题17。如果x = a(cos t + t sin t)，y = a(sin t – t cos t)，则求出$\frac{d^2y}{dx^2}$

解决方案：

According to the question

x = a (cos t + t sin t) …..(1)

y = a (sin t – t cos t) …..(2)

So, \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}   …..(3)

On differentiating eq(1) w.r.t t, we get 

dx/dt = a. $\frac{d}{dt}(cost + t sin t)$

Using chain rule, we get

= a[-sin t +sin t. $\frac{d}{dt}(t)$+ t.$\frac{d}{dt}(sin t)$]

= a [-sin t + sin t + t cos t]

= at cos t 

On differentiating eq(2) w.r.t t, we get 

 dy/dt = a. $\frac{d}{dt}(sin t - t cos t)$

Using chain rule, we get

= a [cos t – [cost.$\frac{d}{dt}(t)$ + t.$\frac{d}{dt}(cos t)$]]

= a[cos t – {cos t – t sin t}]

= at sin t 

Now put the values of dx/dt and dy/dt in eq(1), we get

dy/dx = at sin t/at cos t = tan t

Again differentiating both side w.r.t x, we get 

$\frac{d^2y}{dx^2}= \frac{d}{dx}[\frac{dy}{dx}]$

= $\frac{d}{dx}(tant )$

= sec ² t.$\frac{dt}{dx}$

= sec² t.$\frac{1}{at cos t}$……..[dx/dt = atcost ⇒ dt/dx = 1/atcost]                                      

= sec³t/at 

问题18：如果f(x)= | x |^{如图3所示}，证明对所有实数x都存在f”(x)并找到它。

解决方案：

As we know that |x| = $\begin{cases} x, \hspace{0.2cm}x\geq0\\ -x,\hspace{0.2cm}x<0 \end{cases}$

So, when x ≥ 0, f(x) = |x|³ = x³ 

So, on differentiating both side w.r.t x, we get 

f'(x) = 3x²  

Again, differentiating both side w.r.t x, we get 

f”(x) = 6 x

When x < 0, f(x) = |x|³ = -x³ 

So, on differentiating both side w.r.t x, we get 

f'(x) = – 3x²  

Again, differentiating both side w.r.t x, we get 

f”(x) = -6 x 

So, for f(x) = |x|³, f”(x) exists for all real x, and is given by

f”(x) = $\begin{cases} 6x, \hspace{0.2cm}x\geq0\\ -6x,\hspace{0.2cm}x<0 \end{cases}$

问题19：使用数学归纳法证明$\frac{d}{dx}(x^n)$ =(nx) ^{n – 1}对于所有正整数n。

解决方案：

So, P(n) = $\frac{d}{dx}(x^n)$= (nx)^{n – 1}

For n = 1:

P(1) : $\frac{d}{dx}(x^1)$= (1x)1 ^{– 1} =1

Hence, P(n) is true for n = 1

Let us considered P(k) is true for some positive integer k.

So, P(k): $\frac{d}{dx}(x^k)$= (kx)^{k – 1}

For P(k + 1): $\frac{d}{dx}(x^{k+1})$= ((k + 1)x)^{(k + 1) – 1}   

x ^k $\frac{d}{dx} (x)$+ x.$\frac{d}{dx} (x^k)$ ….(Using applying product rule)

= x ^k .1 + x . k . x ^k-1

= x ^k + k x ^k

= (k + 1) x ^k

= (k + 1) x^{(k + 1) – 1}

Hence, P(k+1) is true whenever P(k) is true.

So, according to the principle of mathematical induction, P(n) is true for every positive integer n.

Hence proved.

问题20.使用sin(A + B)= sin A cos B + cos A sin B的事实和微分，求出余弦的和公式。

解决方案：

According to the question

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

On differentiating both sides w.r.t x, we get

$\frac{d}{dx} [sin(A+B)]$ = $\frac{d}{dx}(sin A cos B)$+ $\frac{d}{dx}(cos A sin B)$ 

⇒ cos (A + B).$\frac{d}{dx}(A+B)$= cos B. $\frac{d}{dx}(sin A)$ + sin A. $\frac{d}{dx}(cos B)$ + sin B.$\frac{d}{dx}(cos A)$+ cos A. $\frac{d}{dx}(sin B)$

⇒ cos (A+B).$\frac{d}{dx}(A+B)$ = cos B.cos A$\frac{dA}{dx}$+ sin A (-sin B) $\frac{dB}{dx}$+ sin B (-sin A).$\frac{dA}{dx}$+ cos A cos B $\frac{dB}{dx}$

⇒ cos (A + B).$[ \frac{dA}{dx}+ \frac{dB}{dx}]$=(cos A cos B – sin A sin B). $[ \frac{dA}{dx}+ \frac{dB}{dx}]$

Hence, cos (A + B) = cos A cos B – sin A sin B

问题21.是否存在一个在任何地方都连续但不能精确地区分为两点的函数?证明你的答案。

解决方案：

Let us consider a function f given as

 f(x) = |x – 1| + |x – 2|

As we already know that the modulus functions are continuous at every point

So, there sum is also continuous at every point but not differentiable at every point x = 0

Let x = 1, 2

Now at x = 1

L.H.D = lim _{x⇢ 1^–} $\frac{f(x) - f(1)}{(x-1)}$

L.H.D = lim_h⇢0 $\frac{f(1-h) - f(1)}{-h}$

= lim_h⇢0 $\frac{|1-h-1| + |1-h-2| - |1-1|-|1-2|}{-h}$

= lim_h⇢0 $\frac{|1-h-1| + |1-h-2| - |0|-|-1|}{-h}$

= lim_h⇢0 $\frac{|-h| + |-h-1| - 1}{-h}$

= lim_h⇢0 $\frac{h - (-h-1) - 1}{-h}$

= lim_h⇢0$\frac {h + h + 1 - 1}{-h}$

= lim_h⇢0$\frac{2h}{-h}$

= -2

R.H.D = lim_x⇢1⁺ $\frac{f(x) - f(1)}{(x-1)}$

R.H.D = lim_h⇢0 $\frac{f(1+h) - f(1)}h$

= lim_h⇢0 $\frac{|1+h-1| + |1+h-2| - |1-1|-|1-2|}{h}$

= lim_h⇢0$\frac{|1+h-1| + |1+h-2| - |0|-|-1|}{h}$

= lim_h⇢0 $\frac{|h| + |h-1| - 1}{h}$

= lim_h⇢0 $\frac{h - (h-1) - 1}{h}$

= lim_h⇢0 $\frac{h - h + 1 - 1}{h}$

= lim_h⇢0 $\frac 0{h}$

= 0

Since L.H.D ≠ R.H.D

So given function f is not differentiable at x = 1.

Similarly, we get that the given function is not differentiable at x = 2.

Hence, there exist a function which is continuous everywhere but not differentiable to exactly two points.

问题22：如果$y=\begin{bmatrix} f(x) & g(x) & h(x)\\ l & m & n\\ a & b & c \end{bmatrix}$ ，证明$\frac{dy}{dx}= \begin{bmatrix} f'(x) & g'(x) & h'(x)\\ l & m & n\\ a & b & c \end{bmatrix}$

解决方案：

Given that$y=\begin{bmatrix} f(x) & g(x) & h(x)\\ l & m & n\\ a & b & c \end{bmatrix}$

⇒ y =(mc – nb) f(x)- (lc – na )g(x) +(lb – ma) h(x)

$\frac{dy}{dx}=\frac{d}{dx}$[(mc -nb) f(x)] – $\frac{d}{dx}$ [(lc – na) g(x)] + $\frac{d}{dx}$[(lb – ma) h(x)]

= (mc – nb) f'(x) – (lc – na) g'(x) + (lb – ma ) h’ (x)

$= \begin{bmatrix} f'(x) & g'(x) & h'(x)\\ l & m & n\\ a & b & c \end{bmatrix}$

So, 

$\frac{dy}{dx}= \begin{bmatrix} f'(x) & g'(x) & h'(x)\\ l & m & n\\ a & b & c \end{bmatrix}$

Hence proved.

问题23：如果y = $e^{a cos^{-1}x}$ ，-1≤x≤1，表明$(1-x^2)\frac{d^2y}{dx^2}- x \frac{dy}{dx}-a^2y=0$

解决方案：

According to the question

y = $e^{a cos^{-1}x}$

Now we are taking logarithm on both sides,

log y = a cos^-1 x log e

log y = a cos ^-1 x 

On differentiating both sides w.r.t x, we get

$\frac{1}{y}\frac{dy}{dx}=a.\frac{-1}{\sqrt{1-x^2}}$

⇒$\frac{dy}{dx}=\frac{-ay}{\sqrt{1-x^2}}$

On squaring both sides,we get

$[\frac{dy}{dx}]^2=\frac{a^2y^2}{1-x^2}$

⇒(1-x ²)$[\frac{dy}{dx}]^2$ =a ² y ²

On differentiating again both the side w.r.t x, we get

$[\frac{dy}{dx}]^2 \frac{d}{dx}(1-x^2)+(1-x^2)\frac{d}{dx} [[\frac{dy}{dx}]^2]=a^2 \frac{d}{dx}(y^2)$

⇒$[\frac{dy}{dx}]^2(-2x)+(1-x^2).2 \frac{dy}{dx}. \frac{d^2y}{dx^2}=a^2.2y \frac{dy}{dx}$

⇒$-x \frac{dy}{dx}+(1-x^2) \frac{d^2y}{dx^2}= a^2.y$

$(1-x^2)\frac{d^2y}{dx^2}- x \frac{dy}{dx}-a^2y=0$

Hence proved