calculo. hoja 3. derivadas...
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Dpto. Matematica Aplicada. E.T.S.A.M. Calculo. Derivadas parciales.
CALCULO. Hoja 3.
Derivadas parciales.
1. Calcular las derivadas parciales de las siguientes funciones:
(a) f(x, y) = x cos x cos y
(b) g(x, y) = (x2 + y2) log(x2 + y2)
(c) h(x, y) = xex2+y2
(d) i(x, y) =x2 + y2
x2 − y2
(e) j(x, y) = exy log(x2 + y2)
(f) k(x, y) = (cos y)exy sin x
Solucion:
(a)∂
∂xf(x, y) = cos x cos y − x sin x cos y
∂
∂yf(x, y) = −x cos x sin y
(b)∂
∂xg(x, y) = 2x ln (x2 + y2) + 2x
∂
∂yg(x, y) = 2y ln (x2 + y2) + 2y
(c)∂
∂xh(x, y) = ex
2+y2 + 2x2ex2+y2
∂
∂yh(x, y) = 2xyex
2+y2
(d)∂
∂xi(x, y) =
−4xy2
(x2 − y2)2
∂
∂yi(x, y) =
4yx2
(x2 − y2)2
(e)∂
∂xj(x, y) = exy
y(ln(x2+y2))x2+y3 ln(x2+y2)+2x
x2+y2
∂
∂yj(x, y) = exy
x3 ln(x2+y2)+x(ln(x2+y2))y2+2y
x2+y2
(f)∂
∂xk(x, y) = (cos y) yexy sinx+ (cos y) exy cosx
∂
∂yk(x, y) = − (sin y) exy sinx+ (cos y)xexy sinx
2. Calcular todas las derivadas primeras y segundas de las siguientes funciones:
(a) f(x, y) =2xy
(x2 + y2)2(x, y) 6= (0, 0)
Dpto. Matematica Aplicada. E.T.S.A.M. Calculo. Derivadas parciales.
(b) g(x, y, z) =ez + 1
x+ xe−yx 6= 0
(c) h(x, y) = cos(xy2)
(d) j(x, y) =1
cos2 x+ e−y
(e) k(x, y) = x
(arctan(
x
y)
)(f) l(x, y) = cos
√x2 + y2
(g) m(x, y) = e−x2−y2
(h) n(x, y) = sin(x2 − 3xy)
(i) p(x, y) = x2y2e2xy
(j) q(x, y) = e−xy2 + y3x4
Solucion:
(a)∂
∂xf(x, y) = −2y 3x2−y2
(x2+y2)3∂
∂yf(x, y) = 2x x2−3y2
(x2+y2)3
∂2
∂x2f(x, y) = 24xy x2−y2
(x2+y2)4∂2
∂y2f(x, y) = −24xy x2−y2
(x2+y2)4
∂2
∂y∂xf(x, y) = −6x4−6x2y2+y4
(x2+y2)4
(b)∂
∂xg(x, y, z) = − ez+1
(1+e−y)x2
∂
∂yg(x, y, z) = ez+1
xe−y
(1+e−y)2∂
∂zg(x, y, z) =
ez
x(1+e−y)
∂2
∂x2g(x, y, z) = 2 ez+1
(1+e−y)x3
∂2
∂y2g(x, y, z) = ez+1
xe−y e−y−1
(1+e−y)3
∂2
∂z2g(x, y, z) = ez
x(1+e−y)
∂2
∂y∂xg(x, y, z) = − ez+1
x2(1+e−y)2e−y
∂2
∂z∂xg(x, y, z) = − ez
(1+e−y)x2
∂2
∂z∂yg(x, y, z) = 1
xez−y
(1+e−y)2
(c)∂
∂xh(x, y) = − (sinxy2) y2
∂
∂yh(x, y) = −2 (sin xy2)xy
∂2
∂x2h(x, y) = − (cosxy2) y4
∂2
∂y2h(x, y) = −4 (cosxy2) x2y2 − 2 (sin xy2)x
∂2
∂y∂xh(x, y) = −2 (cosxy2) y3x− 2 (sin xy2) y
(d)∂
∂xj(x, y) = sin 2x
(cos2 x+e−y)2∂
∂yj(x, y) = e−y
(cos2 x+e−y)2
∂2
∂x2j(x, y) = 2
3 cos2 x−2 cos4 x−e−y+2(cos2 x)e−y
cos6 x+3(cos4 x)e−y+3(cos2 x)e−2y+e−3y
∂2
∂y2j(x, y) = −e−y −e−y+cos2 x
cos6 x+3(cos4 x)e−y+3(cos2 x)e−2y+e−3y :
∂2
∂y∂xj(x, y) = 2e−y sin 2x
(cos2 x+e−y)3
(e)∂
∂xk(x, y) = arctan x
y+ xy
x2+y2∂
∂yk(x, y) = −x2
x2+y2
∂2
∂x2k(x, y) = 2y3
(x2+y2)2∂2
∂y2k(x, y) = 2yx2
(x2+y2)2∂2
∂y∂xk(x, y) = −2xy2
(x2+y2)2
Dpto. Matematica Aplicada. E.T.S.A.M. Calculo. Derivadas parciales.
(f)∂
∂xl(x, y) = − sin
√(y2+x2)√
(y2+x2)x
∂
∂yl(x, y) = − sin
√(y2+x2)√
(y2+x2)y
∂2
∂x2l(x, y) = −
(cos
√(y2+x2)
)x2√
(y2+x2)+(sin√
(y2+x2))y2(√
(y2+x2))3
∂2
∂y2l(x, y) = −
(sin√
(y2+x2))x2+
(cos
√(y2+x2)
)y2√
(y2+x2)(√(y2+x2)
)3
∂2
∂y∂xl(x, y) = −
((cos
√(y2 + x2)
)√(y2 + x2)− sin
√(y2 + x2)
)x y(√
(y2+x2))3
(g)∂
∂xm(x, y) = −2xe−x2−y2 ∂
∂ym(x, y) = −2ye−x2−y2
∂2
∂x2m(x, y) = −2e−x2−y2+4x2e−x2−y2 ∂2
∂y2m(x, y) =−2e−x2−y2+4y2e−x2−y2
∂2
∂y∂xm(x, y) = 4yxe−x2−y2
(h)∂
∂xn(x, y) = 2 (cos (x2 − 3yx))x−3 (cos (x2 − 3yx)) y
∂
∂yn(x, y) =−3 (cos (x2 − 3yx))x
∂2
∂x2n(x, y) = −4 (sin (x2 − 3yx))x2+12 (sin (x2 − 3yx)) yx−9 (sin (x2 − 3yx))
y2 + 2 cos (x2 − 3yx)
∂2
∂y2n(x, y) = −9 (sin (x2 − 3yx))x2
∂2
∂y∂xn(x, y) = 6 (sin (x2 − 3yx))x2 − 9 (sin (x2 − 3yx)) yx− 3 cos (x2 − 3yx)
(i)∂
∂xp(x, y) = 2xy2e2yx + 2x2y3e2yx
∂
∂yp(x, y) = : 2x2ye2yx + 2x3y2e2yx
∂2
∂x2p(x, y) = 2y2e2yx + 8xy3e2yx + 4x2y4e2yx
∂2
∂y2p(x, y) = 2x2e2yx + 8x3ye2yx + 4x4y2e2yx
∂2
∂y∂xp(x, y) = 4xye2yx + 10x2y2e2yx + 4x3y3e2yx
(j)∂
∂xq(x, y) = −y2e−xy2 + 4x3y3
∂
∂yq(x, y) = −2yxe−xy2 + 3x4y2
∂2
∂x2q(x, y) = y4e−xy2 +12x2y3
∂2
∂y2q(x, y) = −2xe−xy2 +4y2x2e−xy2 +6x4y
∂2
∂y∂xq(x, y) = −2ye−xy2 + 2y3xe−xy2 + 12x3y2
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