Job 13 higher derivative

1. Fill in the blanks

(1)

2.

(1) $ (3\cos x +4 \sin x) e^{-2x}$


(2) because
\[
Y ' = 2x \ln (1+x^2)
+ (1+x^2) \frac{2x}{1+x^2}
= 2x\ln (1+x^2) +2x,
\]
So
\[
Y ' =2\ln (1+x^2) + \frac{4 x^2}{1+x^2}+2.
\]




3.

(1) AS
\[
y= \cos^2 x= \frac12 (\cos 2x + 1),
\]
So when $n \geq 1$,
\[
y^{(N)} (x) =2^{n-1} \cos (2x + \frac {n \pi}{2}).
\]

(2) as
\[
Y ' = e^{-x}-X E^{-x},
\]
and
\[
Y ' =-2 e^{-x} +x e^{-x},
\]
Guess
\[
y^{(N)} (x) =
( -1) ^{n-1} (n e^{-x}-X E^{-x}).
\]
The following mathematical induction is used to prove that when the $n =1$ is apparently established, assuming $n =k$, then
\[
y^{(K+1)} (x)
= (y^{(k)} (x)) '
= ( -1) ^{k-1} (ke^{-x}-x e^{-x}) '
= ( -1) ^{k-1} (-ke^{-x}-e^{-x}+xe^{-x})
= ( -1) ^{k} ((k+1) e^{-x}-xe^{-x}),
\]
Launched when $n =k+1$ is also established, so for any $n \geq 1$,
\[
y^{(N)} (x) =
( -1) ^{n-1} (n e^{-x}-X E^{-x}).
\]




4. Process a little, just give the answer

(1)
\[
Y ' =-\frac{1}{\sin^2 (x+y)},
\qquad
Y ' =-2 \frac{\cos^3 (x+y)}
{\sin^5 (X+y)}.
\]


(2) on both sides to take the logarithm will be relatively good, or calculate the first derivative, the use of primitive simplification, get
\[
Y ' =\frac{Y-e^{x+y}}
{E^{x+y}-x}
=\frac{y-1-xy}{1+xy-x},
\]
\[
Y ' '
=\frac{\frac{y-1-xy}{1+xy-x}-y-x\frac{y-1-xy}{1+xy-x}}{1+xy-x}
-\frac{(Y-1-xy) (y+x\frac{y-1-xy}{1+xy-x}-1)} {(1+xy-x) ^2}.
\]
Really don't want to simplify ...




5. The answer, the calculation process is slightly

(1)
\[
\FRAC{DY}{DX}
=\FRAC{\FRAC{DY}{DT}}
{\FRAC{DX}{DT}}= t\sin T,
\qquad
\FRAC{D^2Y}{DX^2}
=\FRAC{\FRAC{D}{DT} (\frac{dy}{dx})}
{\FRAC{DX}{DT}}
= (\sin t +t \cos t) \tan T.
\]



(2)
\[
\FRAC{DY}{DX}
=\FRAC{\FRAC{DY}{DT}}
{\FRAC{DX}{DT}}= T,
\qquad
\FRAC{D^2Y}{DX^2}
=\FRAC{\FRAC{D}{DT} (\frac{dy}{dx})}
{\FRAC{DX}{DT}}
=\frac{1}{f ' (t)}.
\]


6. Solution: First $f (x) $ at 0 points in succession, so
\[
\lim_{x\to 0^-} f (x)
= \lim_{x\to 0^-} e^x = f (0)
= C,
\]
That is $c =1$. Again because $f (x) $ in the 0.1-derivative existence, then $f ' _+ (0) =f_-' (0) $, due
\[
F_-' (0) = \lim_{\delta x\to 0^-}
\frac{E^{\delta x}-f (0)} {\delta x}=
\lim_{\delta x\to 0^-}
\frac{E^{\delta x}-1} {\delta x}=1
\]
and
\[
F ' _+ (0) =\lim_{\delta x\to 0^+}
\frac{A (\delta x) ^2 +b \delta x +1-f (0)} {\delta X}
=\lim_{\delta x\to 0^+}
\frac{A (\delta x) ^2 +b \delta x} {\delta x}
=b,
\]
So $b =1$. According to the above discussion, $f ' (0) =1$, then
\[
F ' (x) =
\begin{cases}
E^x, & x<0, \ \
1, & x=0, \ \
2a x +1, & x>0.
\end{cases}
\]
Obviously, $f ' (x) $ continuous, and according to the title, Get $f ' _+ (0) =f_-' (0) $, then
\[
F_-"(0) = \lim_{\delta x\to 0^-}
\frac{e^{\delta x}-f ' (0)} {\delta x}=1,
\]
and
\[
F ' _+ (0) =\lim_{\delta x\to 0^+}
\frac{2a \delta x +1-f ' (0)} {\delta x}=2a,
\]
Is $a =\frac12$, so
\[
A=\FRAC12, B=1, C=1.
\]


Note: The value of $f ' (0) $ cannot be directly obtained by the derivation of $f (x) $, or by definition, because it is a staging point.

Job 13 higher derivative