CalculusEasyMade_CH12

1

x, y=symbols('x y', real=True)
eq=x**3/3-2*x**3*y-2*y**2*x+y/3
eq
$- 2 x^{3} y + \frac{x^{3}}{3} - 2 x y^{2} + \frac{y}{3}$

deqdx=diff(eq, x)
deqdx
$- 6 x^{2} y + x^{2} - 2 y^{2}$

deqdy=diff(eq, y)
deqdy
$- 2 x^{3} - 4 x y + \frac{1}{3}$

3

$r^{2}= \left(- a + x\right)^{2} + \left(- b + y\right)^{2} +\left(- c + z\right)^{2}$

a, b, c, r, x, y, z=symbols('a b c r x y z', real=True)
eq=r**2-((x-a)**2+(y-b)**2+(z-c)**2)
eq
$r^{2} - \left(- a + x\right)^{2} - \left(- b + y\right)^{2} - \left(- c + z\right)^{2}$

deqdx=idiff(eq, r, x)
deqdx
$\frac{- a + x}{r}$

deqdy=idiff(eq, r, y)
deqdy
$\frac{- b+y}{r}$

deqdz=idiff(eq, r, z)
deqdz
$\frac{- c+z}{r}$

deqdx+deqdy+deqdz
$\frac{- a + x}{r} + \frac{- b + y}{r} + \frac{- c + z}{r}$

diff(deqdx, x)+diff(deqdy, y)+diff(deqdz, z)
$\frac{3}{r}$

5_a

u, v=symbols('u v', real=True)
y=u**3*sin(v)
y
u³sin(v)

dydu=diff(y, u)
dydu
3u²sin(v)

dydv=diff(y, v)
dydv
u³cos(v)

dy=dydu+dydv
dy
u³cos(v)+3u²sin(v)

위 결과를 보다 자세히 나타내면 다음과 같습니다.
u³cos(v)⋅⋅du+3u²sin(v)⋅dv

5_b

x, u=symbols('x u', real=True)
y=(sin(x))**u
y
$ \sin^{u}{\left(x \right)}$

dydu=diff(y, u)
dydu
log(sin(x))sin^u(x)

dydx=diff(y, x)
dydx
$\frac{u \sin^{u}{\left(x \right)} \cos{\left(x \right)}}{\sin{\left(x \right)}}$

dy=dydx+dydu
dy
$\frac{u \sin^{u}{\left(x \right)} \cos{\left(x \right)}}{\sin{\left(x \right)}} + \log{\left(\sin{\left(x \right)} \right)} \sin^{u}{\left(x \right)}$

5_c

u, v=symbols('u v', real=True)
y=log(u)/v
y
$\frac{log(u)}{v}$

dydu=diff(y, u)
dydu
$\frac{1}{uv}$

dydv=diff(y, v)
dydv
$−\frac{log(u)}{v^2}$

dy=dydu+dydv
dy
$ - \frac{\log{\left(u \right)}}{v^{2}} + \frac{1}{u v}$

7

x, y=symbols('x y', real=True)
u=x+2*x*y+y
u
2xy+x+y

dudx=diff(u, x)
dudx
2y+1

sol_x=solve(dudx)
sol_x
[-1/2]

dudy=diff(u, y)
dudy
2x+1

u.subs({x:-1/2, y:-1/2})
−0.5

u.subs({x:0, y:0})
0

u.subs({x:-1, y:-1})
0

위 결과와 같이 (x, y)=($-\frac{1}[2}, \frac{1}[2})에서 극소값 -0.5를 갖습니다.

10

x, y=symbols('x y', real=True)
u=exp(x+y)/(x*y)
u
$ \frac{e^{x + y}}{x y}$
dudx=diff(u, x)
dudx
$\frac{e^{x + y}}{x y} - \frac{e^{x + y}}{x^{2} y}$
sol_x=solve(dudx, x)
sol_x
[1]
diff(dudx, x).subs(x, 1)
$\frac{e^{y + 1}}{y}$
dudy=diff(u, y)
dudy
$\frac{e^{x + y}}{x y} - \frac{e^{x + y}}{x y^{2}}$
sol_y=solve(dudy, y)
sol_y
[1]
diff(dudy, y).subs(y, 1)
$\frac{e^{x + 1}}{x}$

x=1, y=1 에서 극값이 존재하면 x,y의 부호에 따라 극대값 또는 극소값이 결정됩니다.