Calculo diferencial clase 5-derivadas de funciones trigonometricas
CALCULO. Hoja 3. Derivadas...
Transcript of CALCULO. Hoja 3. Derivadas...
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