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