Chapter two self-test questions

1. Fill in the blanks


(1) $\frac12 F ' (X_0) $


To make $x =x_0+ \delta x$, the
\[
\mbox{native}=\lim_{\delta x\to 0}
\frac{F (x_0 +\frac{\delta x}{2})-F (x_0)} {\delta X}
=\frac12\lim_{\delta x\to 0}
\frac{F (x_0 +\frac{\delta x}{2})-F (x_0)} {\frac{\delta x}{2}}=\frac12 F ' (X_0).
\]

If not strictly utilized on $x $ Rockwell can be seen more intuitively
\[
\mbox{Native}=\lim_{x\to X_0}
\frac{\frac12 F ' (\frac{x+x_0}{2})} {1}
=\frac12 F ' (X_0).
\]


(2)-4

Because
\[
\frac{2}{x^3} f ' (\frac{1}{x^2}) =\frac1x,
\]
So
\[
F ' (\frac{1}{x^2}) =-2x^2,
\]
So that the $x =\sqrt 2$ can be.


(3) $ ( -1) ^{n+1} n!$

Because
\[
f^{(N)} (x) = \frac{n!} {(1-x) ^{n+1}}
\]


(4)
\[
Y ' = \frac{e^x+2x \cos (x^2+y^2)-y^2}
{2xy-2y \cos (x^2 +y^2)}
\]



2. (1)
\[
\begin{aligned}
Y ' &=\left (
\arcsin \frac{2\sin x+1}{2+\sin x}
\right) '
\\
&=\left (
\frac{2\sin x+1}{2+\sin x}
\right) '
\frac{1}{\sqrt{1-\left (
\frac{2\sin x+1}{2+\sin x}
\right) ^2}}
\\
&=
\left (2-
\frac{3}{2+\sin X}
\right) '
\frac{1}{\sqrt{1-\left (
\frac{2\sin x+1}{2+\sin x}
\right) ^2}}
\\
&= \frac{3 \cos x}{(2+\sin x) ^2}
\frac{|2+\sin x|} {\SQRT 3 |\cos x|}
\\
&=
\frac{3 \cos x}{(2+\sin x) ^2}
\frac{2+\sin x} {\SQRT 3 \cos x} \qquad (|x|<\frac\pi 2)
\\
&= \frac{\sqrt 3}{2+\sin X}
\end{aligned}
\]
So
\[
DY = \frac{\sqrt 3}{2+\sin x} dx.
\]


(2) as
\[
Y ' = (f (x+y)) '
= F ' (x+y) (1+y '),
\]
So
\[
Y ' = \frac{f ' (X+y)} {1-f ' (x+y)}.
\]
So
\[
\begin{aligned}
Y ' &= \left (
\frac{F ' (X+y)} {1-f ' (x+y)}
\right) '
\\
&= \frac{F ' (x+y) (1+y ') (1-f ' (x+y)) +
F ' (x+y) F ' (x+y) (1+y ')
}
{(1-f ' (x+y)) ^2}
\\
&=
\frac{F ' (x+y) (1+y ')
}
{(1-f ' (x+y)) ^2}
\\
&= \frac{F ' (X+y)} {(1-f ' (x+y)) ^3}.
\end{aligned}
\]


(3) First
\[
\frac{dx}{dt}= 6t +2,
\qquad \frac{dy}{dt}=
\frac{e^y \cos T} {1-e^{y}\sin T}
=\frac{e^y \cos t} {2-y},
\]
So
\[
\FRAC{DY}{DX}
= \frac{dy}{dt}/\frac{dx}{dt}
=\frac{e^y \cos T} {(2-y) (6t+2)},
\]
That
\[
\FRAC{D}{DT} (\frac{dy}{dx})
= \frac{e^y y ' (t) \cos t-e^y \sin T} {(2-y) (6t+2)}-\frac{e^y \cos t [6 (2-Y)-y ' (6t+2)]} {(2-y) ^2 (6t+2) ^2}
\]
And because when $t =0$, $x =3,y=1$ so
\[
\frac{d^2 y}{d x^2} \bigg|_{t=0}
=\FRAC12 e^2-\frac34 E.
\]


3. Proof: According to the question has
\[
\lim_{x\to 0} \exp
\left (
\frac{\ln (1+x +\frac{f (x)}{x})}{x}
\right)
=\exp
\left (\lim_{x\to 0}
\frac{\ln (1+x +\frac{f (x)}{x})}{x}
\right)
=e^3,
\]
According to the monotonicity of exponential function, there are
\[
\lim_{x\to 0}
\frac{\ln (1+x +\frac{f (x)}{x})}{x}=3.
\tag{*}
\]
So
\[
\lim_{x\to 0}
{\ln (1+x +\frac{f (x)}{x})}
=
\lim_{x\to 0}
\frac{\ln (1+x +\frac{f (x)}{x})}{x}
\cdot \lim_{x\to 0} x=0=\ln 1.
\]
According to the monotonicity of the logarithmic function, there are
\[
\lim_{x\to 0} (x +\frac{f (x)}{x}) =0.
\tag{**}
\]
First of all
\[
\lim_{x\to 0} (X^2+{f (x)}) =\lim_{x\to 0} (x +\frac{f (x)}{x}) \cdot \lim_{x\to 0} x=0,
\]
Launch
\[
F (0) =\lim_{x\to 0} f (x) =-\lim_{x\to 0}x^2=0.
\]
Second, by $ (* *) $
\[
\lim_{x\to 0} \frac{f (x)}{x}
= \lim_{x\to 0} \frac{f ' (x)}{1}=f ' (0) =0.
\]
Again, by $ (*) $ and $ (* *) $
\[
\lim_{x\to 0}
\frac{\ln (1+x +\frac{f (x)}{x})}{x}
=
\lim_{x\to 0}
\frac{x +\frac{f (x)}{x}}{x}
=
\lim_{x\to 0}
\frac{x^2 +{f (x)}}{x^2}
= 1+\frac12 F ' (0)
=3
\]
That is $f ' (0) =4$. Sum up
\[
F (0) =0,f ' (0) =0,f "(0) =4.
\]

(2)
\[
\mbox{native}=\lim_{x\to 0} \exp \frac{\ln (1+\frac{f (x)}{x})}
{x}
= \exp \left (
\lim_{x\to 0} \frac{\ln (1+\frac{f (x)}{x})}
{x}\right)
=\exp \left (
\lim_{x\to 0} \frac{f (x)}{x^2}\right)
=e^2.
\]



(4) First, assuming $x =0$, the
\[
F (0) = 0,
\]
Second, assume that $x >0$, because $e ^{tx}\to +\infty$ when $t \to +\infty$, the
\[
F (x) =\lim_{t\to +\infty}
\frac{x} {2+X^2-E^{TX}}=0,
\]
Finally, when $x <0$, because $e ^{tx}\to 0$ when $t \to +\infty$, the
\[
F (x) =\lim_{t\to +\infty}
\frac{x} {2+X^2-E^{TX}}=\frac{x}{2+x^2}.
\]
Sum up
\[
f (x) =
\begin{cases}
0, & X\geq 0,
\\
\frac{x}{2+x^2}, & x< 0.
\end{cases}
\]
Notice that $f (x) $ is a continuous function, and when $x \neq 0$ can be directed. and
\[
F ' _+ (0) = 0, \qquad
F_-' (0) =\lim_{\delta x\to 0^-}
\frac{\frac{\delta x}{2 + (\delta x) ^2}-0} {\delta x}=\FRAC12,
\]
therefore $f (x) $ is not =0$ at the $x.



5. According to the Leibniz formula
\[
\begin{aligned}
F^{(2001)} (x) &= \frac12 (x^2 \sin (2x)) ^{(2001)}
\\
&AMP;=\FRAC12 x^2 \sin^{(2001)} (2x)
+ c_{2001}^1 x \sin^{(2x)}
+ c_{2001}^2 \sin^{(1999)} (2x).
\end{aligned}
\]
Notice that the $x =0$, the first two items are zero, so
\[
F^{(2001)} (0)
= c_{2001}^2 2^{1999} \sin (2\cdot 0 +\frac{1999\pi}{2})
=c_{2001}^2 2^{1999} \sin (499 \cdot 2\pi +\frac{3\pi}{2})
=-c_{2001}^2 2^{1999}.
\]



6. Proof: According to the question has
\[
F (0+0) =f (0) F (0),
\]
Therefore $f (0) =1$. and
\[
F ' (x)
= \lim_{\delta x\to 0}
\frac{f (X+\delta x)-F (x)} {\delta X}
=
\lim_{\delta x\to 0}
\frac{f (x) F (\delta X)-F (x)} {\delta X}
= f (x) \lim_{\delta x\to 0}
\frac{f (\delta x)-1} {\delta X}
=f (x) F ' (0) =f (x).
\]

Chapter two self-test questions