Chapter II Self-Test

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 $ (* *) $
\[
\lim_{x\to 0}
\frac{\ln (1+x +\frac{f (x)}{x})}{x}
=
\lim_{x\to 0}
\frac{x +\frac{f (x)}{x}}{x}
=
\lim_{x\to 0}
\frac{x^2 +{f (x)}}{x^2}
=1+\lim_{x\to 0}
\frac{f ' (x)}{2x}
=1+\lim_{x\to 0}
\frac{f ' (x)-F ' (0)}{2x}
= 1+\frac12 F ' (0)
=3
\]
That is $f ' (0) =4$. Sum up
\[
F (0) =0,f ' (0) =0,f "(0) =4.
\]

Note: The front is only used once indeterminate form, the last step is defined, because according to the conditions, the second derivative is not continuous, if the use of indeterminate form no way to get the limit, so with the definition, mathematics is very subtle ...

(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 II Self-Test