#import "../template.typ": *

#show: project.with(
  title: "Calculus Homework #6",
  authors: ((
    name: "<NAME> (#47)",
    email: "<EMAIL>",
    phone: "3230104585"
  ),),
  date: "October 31, 2023",
)

= P98 习题2-1 12
#prob[求导数：$display(y = 1/x^2 + root(3,x^2) + 1/root(3,x) + root(3,2))$]
$ y'
= (x^(-2) + x^(2/3) + x^(-1/3) + root(3,2))'
= -2 x^(-3) + 2/3 x^(-1/3) - 1/3 x^(-4/3)
= - 2/(x^3) + 2/(3 root(3,x)) - 1/(3 x root(3,x))
$

= P98 习题2-1 15
#prob[求导数：$display(y = x^2 sin x + 2 x sin x - cos pi / 5)$]
$ y'
= 2x sin x + x^2 cos x + 2 sin x + 2x cos x
= 2 (x+1) sin x + x(x+2)cos x
$

= P99 习题2-1 18
#prob[求导数：$display(y = sqrt(x)/(sqrt(x)+1))$]
$ y'
= ((sqrt(x))' (sqrt(x) + 1) - (sqrt(x) + 1)' sqrt(x))/((sqrt(x) + 1)^2)
= (1/(2 sqrt(x)) (sqrt(x) + 1) - 1/(2 sqrt(x)) sqrt(x))/(x + 2 sqrt(x) + 1)
= 1/(2 x sqrt(x) + 4 x + 2 sqrt(x))
$

= P99 习题2-1 21
#prob[求导数：$display(y = 1/(1+sqrt(x)) - 1/(1-sqrt(x)))$]
$ y'
&= frac(-(1+sqrt(x))', (1+sqrt(x))^2) - frac(-(1-sqrt(x))', (1-sqrt(x))^2)
= -1/(2 sqrt(x)) (1/(1+sqrt(x))^2 + 1/(1-sqrt(x))^2)\
&= - frac((1+sqrt(x))^2 + (1-sqrt(x))^2, 2 (1-x)^2 sqrt(x))
= - frac(x+1, (1-x)^2 sqrt(x))
$

= P99 习题2-1 24
#prob[求导数：$display(y = frac(sin x - cos x, sin x + cos x))$]
$ y'
& = frac(
	(sin x - cos x)' (sin x + cos x) - (sin x + cos x)' (sin x - cos x),
	(sin x + cos x)^2)\
&= frac(
	(cos x + sin x) (sin x + cos x) - (cos x - sin x) (sin x - cos x),
	sin^2 x + 2 sin x cos x + cos^2 x)\
&= 2/(1 + sin 2x)
$

= P99 习题2-1 27
#prob[求导数：$display(y = (x^3 - 1/x^3 + 3)^4)$]
$ y'
= 4 (x^3 - 1/x^3 + 3)^4 (x^3 - 1/x^3 + 3)'
= 12 (x^3 - 1/x^3 + 3)^4 (x^2 + 1/x^4)
$

= P99 习题2-1 30
#prob[求导数：$display(y = (3-2 sin x)^5)$]
$ y'
= 5 (3 - 2 sin x)^4 (3 - 2 sin x)'
= -10 cos x (3 - 2 sin x)^4
$

= P99 习题2-1 33
#prob[求导数：$display(y = sqrt(sin x))$]
$ y' = (cos x)/(2 sqrt(sin x)) $

= P99 习题2-1 36
#prob[求导数：$display(y = sin^2 x/2 + cos^2 x/2)$]
$ y'
= 2 sin x/2 (sin x/2)' + 2 cos x/2 (cos x/2)'
= 2 sin x/2 (1/2 cos x/2) + 2 cos x/2 (-1/2 sin x/2)
= 0
$

= P99 习题2-1 39
#prob[求导数：$display(y = (frac(1 - cos x, 1 + cos x))^3)$]
$ y'
&= 3(frac(1-cos x,1+cos x))^2 (frac(1-cos x,1+cos x))'\
&= 3(frac(1-cos x,1+cos x))^2 frac((1-cos x)'(1+cos x)-(1+cos x)'(1-cos x),(1+cos x)^2)\
&= frac(6 sin x (1-cos x)^2,(1+cos x)^4)
$

= P99 习题2-1 42
#prob[求导数：$display(y = x sec^2 x - tan 2x)$]

$ y'
= sec^2 x + x (2 sec x) (sec x)' - 2 sec 2x
= sec^2 x + 2x sec^2 x tan x - 2 sec 2x
$

= P99 习题2-1 45
#prob[求导数：$display(y = sqrt(x sin 2x))$]
$ y'
= 1/(2 sqrt(x sin 2x)) (x sin 2x)'
= (sin 2x + 2 x cos 2x)/(2 sqrt(x sin 2x))
$

= P99 习题2-1 48
#prob[求导数：$display(y = e^(2x + 3))$]
$ y'
= e^(2x + 3) (2x + 3)'
= 2 e^(2x + 3)
$

= P99 习题2-1 51
#prob[求导数：$display(y = cos sqrt(x) + sqrt(cos x) + sqrt(cos sqrt(x)))$]
$ y'
&= - sin sqrt(x) (sqrt(x))' + 1/(2 sqrt(cos x)) (cos x)' + 1/(2 sqrt(cos sqrt(x))) (cos sqrt(x))'\
&= -(sin sqrt(x))/(2 sqrt(x)) + (sin x)/(2 sqrt(cos x)) - (sin sqrt(x))/(4 sqrt(x cos sqrt(x)))\
$

= P99 习题2-1 54
#prob[求导数：$display(y = ln (sec x + tan x))$]

$ y'
= (sec x + tan x)'/(sec x + tan x)
= (sec x tan x + tan x) / (sec x + tan x)
= (sec x + 1) / (csc x + 1)
$

= P99 习题2-1 57
#prob[求导数：$display(y = sqrt(1 + (ln x)^2))$]

$ y'
= ((1 + (ln x)^2)')/(2 sqrt(1 + (ln x)^2))
= (2 ln x dot 1/x)/(2 sqrt(1 + (ln x)^2))
= (ln x)/(x sqrt(1 + (ln x)^2))
$

= P99 习题2-1 60
#prob[求导数：$display(y = arcsin 1/x)$]

$ y'
= ((1/x)')/sqrt(1 + (1/x)^2)
= (-1/x^2)/sqrt(1 + 1/(x^2))
= -1/(x sqrt(x + 1))
$

= P99 习题2-1 63
#prob[求导数：$display(y = e^x sqrt(1 - e^(2x)) + s e^x)$]

$ y'
&= e^x (sqrt(1 - e^(2x)) + ((1-e^(2x))')/(2 sqrt(1 - e^(2x)))) + ((e^(2x))')/sqrt(1-e^(2x))\
&= (e^x (1 - e^(2x)) - e^(2x) + 2 e^(2x))/sqrt(1-e^(2x))\
&= (e^x + e^(2x) - e^(3x))/sqrt(1-e^(2x))
$

= P100 习题2-1 66
#prob[求导数：$display(y = e^sqrt(ln x))$]

$ y'
&= e^(sqrt(ln x)) (sqrt(ln x))'
= e^(sqrt(ln x)) ((ln x)')/(2 sqrt(ln x))
= (e^sqrt(ln x))/(2x sqrt(ln x))
$

= P100 习题2-1 69
#prob[求导数：$display(y = x^x + x^(1/x))$]

$
y &= e^(x ln x) + e^(1/x ln x)\ => y'
&= e^(x ln x) (x ln x)' + e^(1/x ln x) (1/x ln x)'\
&= x^x (1 + ln x) + x^(1/x) (-1/x^2 ln x + 1/x^2)\
&= x^x (1 + ln x) + x^(1/x - 2) (1 - ln x)
$

= P100 习题2-1 72
#prob[设 $display(y = arctan phi(x)/psi(x))$，求 $display(dy/dx)$．]

$ y'
= display((phi(x)/psi(x))')/(1 + display((phi(x)/psi(x))^2))
= (phi'(x) psi(x) - psi'(x) phi(x))/(phi^2(x) + psi^2(x))
$

= P100 习题2-1 81(1)
#prob[求 $display(y = sin x)$ 在点 $display(x = pi/4)$ 的切线方程与法线方程．]

$ dy/dx = cos x $

所以在点 $x=display(pi/4)$ 的

- 切线方程为：$display(y-sqrt(2)/2 = sqrt(2)/2 (x - pi/4) => y = sqrt(2)/2 x + sqrt(2)/2 - (pi sqrt(2))/8)$；

- 法线方程为：$display(y-sqrt(2)/2 = -sqrt(2) (x - pi/4) => y = -sqrt(2) x + sqrt(2)/2 + (pi sqrt(2))/4)$．

= P100 习题2-1 81(3)
#prob[求 $display(e^(x y) - x^2 + y^3 = 0)$ 在点 $display(x = 0)$ 的切线方程与法线方程．]

$
=> (dif e^(x y))/dx - (dif x^2)/dx + (dif y^3)/dx = 0
=> e^(x y) (y + x dy/dx)- 2x + 3 y^2 dy/dx = 0
$

当 $x=0$ 时，$y=-1$，此时 $display(dy/dx = 1/3)$．

故原函数在点 $x=0$ 的

- 切线方程为：$display(y+1 = 1/3(x - 0) => y = x/3 - 1)$；

- 法线方程为：$display(y+1 = -3(x - 0) => y = -3x - 1)$．

= P100 习题2-1 82(1)
#prob[求导数：$display(y = |x^3|)$]

当 $x>=0$ 时，$y' =(x^3)' = 3 x^2$；当 $x<=0$ 时，$y' = (- x^3)' = -3 x^2$．

注意到当 $x=0$ 时，$y'_+ = y'_- = 0$，故原函数导数在 $RR$ 上存在，且为：

$ y' = cases(
  3 x^2 \,quad x>=0,
	-3 x^2 \, quad x<0
) $

= P100 习题2-1 82(2)
#prob[求导数：$display(y = x |x(x-1)|)$]

当 $x>=1$ 或 $x<=0$ 时，$y' = (x^2 (x-1))' = 2x (x-1) + x^2 = 3x^2 - 2x$；

当 $0<=x<=1$ 时，$y' = (-x^2 (x-1))' = 2x - 3x^2$．

注意到 $x=1$ 时 $y'_+ != y'_-$，故原函数 $y$ 在点 $x=1$ 处不可导．在其余值域上的导数为：

$ y' = cases(
	3x^2 - 2x \, quad x>1,
	2x - 3x^2 \, quad 0<=x<1,
	3x^2 - 2x \, quad x<0
) $

= P100 习题2-1 83(1)
#prob[求导数：$display(f(x) = cases(
	x^3\,quad& x>0,
	x^2\,quad& x<=0,
))$]

当 $x>0$ 时，$f'(x) = 3x^2$；当 $x<=0$ 时 $f'(x) = 2x$．

注意到 $x=1$ 时，$f'_+(x) = f'_-(x) = 0$，故原函数导数在 $RR$ 上存在，且为：

$ f'(x) = cases(
	3x^2 \,quad x>0,
	2x \,quad x<=0
) $

= P100 习题2-1 83(2)
#prob[求导数：$display(f(x) = cases(
	display(x^2 cos 1/x)\,quad& x!=0,
	0\,quad& x=0,
))$]

当 $x!=0$ 时，$display(f'(x) = 2x cos(1/x) + x^2 (-sin(1/x)) (-1/x^2) = 2x cos(1/x) + sin(1/x))$．

另有 $display(lim_(x->0) f'(x) = lim_(x->0) sin(1/x))$ 不存在，故 $f(x)$ 在点 $x=0$ 处不可导．

故原函数 $f(x)$ 的导数为 $f'(x) = display(2x cos (1/x) + sin (1/x)) quad (x!=0)$．