Dpto. Matemática Aplicada. E.T.S.A.M. Cálculo. Derivadas parciales. CÁLCULO. 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 + y 2 ) log(x2 + y 2 ) (c) h(x, y) = xex 2 +y 2 x2 + y 2 x2 − y 2 (e) j(x, y) = exy log(x2 + y 2 ) (d) i(x, y) = (f) k(x, y) = (cos y)exy sin x Solución: (a) (b) (c) (d) (e) (f) ∂ f (x, y) = cos x cos y − x sin x cos y ∂x ∂ f (x, y) = −x cos x sin y ∂y ∂ g(x, y) = 2x ln (x2 + y 2 ) + 2x ∂x ∂ g(x, y) = 2y ln (x2 + y 2 ) + 2y ∂y ∂ 2 2 2 2 h(x, y) = ex +y + 2x2 ex +y ∂x ∂ 2 2 h(x, y) = 2xyex +y ∂y ∂ −4xy 2 i(x, y) = 2 ∂x (x − y 2 )2 ∂ 4yx2 i(x, y) = 2 ∂y (x − y 2 )2 ∂ y (ln(x2 +y 2 ))x2 +y 3 ln(x2 +y 2 )+2x j(x, y) = exy x2 +y 2 ∂x 3 2 2 ∂ x ln(x +y )+x(ln(x2 +y 2 ))y 2 +2y j(x, y) = exy x2 +y 2 ∂y ∂ k(x, y) = (cos y) yexy sin x + (cos y) exy cos x ∂x ∂ k(x, y) = − (sin y) exy sin x + (cos y) xexy sin x ∂y 2. Calcular todas las derivadas primeras y segundas de las siguientes funciones: (a) f (x, y) = 2xy (x2 + y 2 )2 (x, y) 6= (0, 0) Dpto. Matemática Aplicada. E.T.S.A.M. (b) (c) (d) (e) (f) Cálculo. Derivadas parciales. ez + 1 g(x, y, z) = x 6= 0 x + xe−y h(x, y) = cos(xy 2 ) 1 j(x, y) = 2 cos x + e−y x k(x, y) = x arctan( ) y p l(x, y) = cos x2 + y 2 (g) m(x, y) = e−x 2 −y 2 (h) n(x, y) = sin(x2 − 3xy) (i) p(x, y) = x2 y 2 e2xy (j) q(x, y) = e−xy + y 3 x4 2 Solución: ∂ ∂ 2 −y 2 2 2 f (x, y) = −2y (x3x2 +y f (x, y) = 2x (xx2 −3y 2 )3 2 )3 +y ∂x ∂y 2 ∂2 ∂ 2 −y 2 x2 −y 2 f (x, y) = 24xy f (x, y) = −24xy (xx2 +y 4 2 2 2 )4 2 2 (x +y ) ∂x ∂y ∂2 4 −6x2 y 2 +y 4 f (x, y) = −6 x (x 2 +y 2 )4 ∂y∂x ∂ ∂ z ez +1 e−y (b) g(x, y, z) = − (1+e g(x, y, z) = e x+1 (1+e −y )x2 −y )2 ∂x z ∂y (a) e x(1+e−y ) 2 ∂ g(x, y, z) = ∂z ∂ ∂2 z ez +1 e−y −1 g(x, y, z) = 2 g(x, y, z) = e x+1 e−y (1+e −y )3 (1+e−y )x3 ∂x2 ∂y 2 ∂2 ∂2 ez ez +1 −y g(x, y, z) = g(x, y, z) = − x2 (1+e −y )2 e x(1+e−y ) 2 ∂z ∂y∂x ∂2 ∂2 z ez−y g(x, y, z) = − (1+ee−y )x2 g(x, y, z) = x1 (1+e −y )2 ∂z∂x ∂z∂y ∂ ∂ (c) h(x, y) = − (sin xy 2 ) y 2 h(x, y) = −2 (sin xy 2 ) xy ∂x ∂y ∂2 ∂2 2 4 h(x, y) = − (cos xy ) y h(x, y) = −4 (cos xy 2 ) x2 y 2 − 2 (sin xy 2 ) x ∂x2 ∂y 2 ∂2 h(x, y) = −2 (cos xy 2 ) y 3 x − 2 (sin xy 2 ) y ∂y∂x ∂ ∂ −y 2x j(x, y) = (cos2sin j(x, y) = (cos2 ex+e−y )2 (d) x+e−y )2 ∂x ∂y 2 2 4 ∂ 3 cos x−2 cos x−e−y +2(cos2 x)e−y j(x, y) = 2 cos6 x+3(cos4 x)e−y +3(cos2 x)e−2y +e−3y ∂x2 2 ∂ −e−y +cos2 x j(x, y) = −e−y cos6 x+3(cos4 x)e −y +3(cos2 x)e−2y +e−3y : 2 ∂y ∂2 2e−y sin 2x j(x, y) = (cos 2 x+e−y )3 ∂y∂x ∂ ∂ 2 k(x, y) = arctan xy + x2xy k(x, y) = x−x (e) 2 +y 2 +y 2 ∂x ∂y 2 2 ∂ ∂2 ∂ 2 2y 3 2yx2 k(x, y) = k(x, y) = k(x, y) = (x−2xy 2 2 2 2 2 2 2 +y 2 )2 (x +y ) (x +y ) 2 2 ∂x ∂y ∂y∂x Dpto. Matemática Aplicada. E.T.S.A.M. Cálculo. Derivadas parciales. √ √ ∂ ∂ sin (y 2 +x2 ) sin (y 2 +x2 ) √ √ (f) l(x, y) = − x l(x, y) = − y (y 2 +x2 ) (y 2 +x2 ) ∂x ∂y √ √ √ cos (y 2 +x2 ) x2 (y 2 +x2 )+ sin (y 2 +x2 ) y 2 ∂2 √ 3 l(x, y) = − ∂x2 (y 2 +x2 ) √ √ √ sin (y 2 +x2 ) x2 + cos (y 2 +x2 ) y 2 (y 2 +x2 ) ∂2 √ 3 l(x, y) = − ∂y 2 (y 2 +x2 ) p p p ∂2 l(x, y) = − cos (y 2 + x2 ) (y 2 + x2 ) − sin (y 2 + x2 ) x √ y 3 ∂y∂x (y 2 +x2 ) ∂ ∂ 2 2 2 2 m(x, y) = −2xe−x −y m(x, y) = −2ye−x −y ∂x ∂y 2 ∂2 ∂ 2 2 2 2 −x2 −y 2 2 −x2 −y 2 m(x, y) = −2e +4x e m(x, y) = −2e−x −y +4y 2 e−x −y 2 2 ∂x ∂y 2 ∂ 2 2 m(x, y) = 4yxe−x −y ∂y∂x ∂ ∂ (h) n(x, y) = 2 (cos (x2 − 3yx)) x−3 (cos (x2 − 3yx)) y n(x, y) = −3 (cos (x2 − 3yx)) x ∂x ∂y ∂2 n(x, y) = −4 (sin (x2 − 3yx)) x2 + 12 (sin (x2 − 3yx)) yx − 9 (sin (x2 − 3yx)) ∂x2 y 2 + 2 cos (x2 − 3yx) ∂2 n(x, y) = −9 (sin (x2 − 3yx)) x2 2 ∂y ∂2 n(x, y) = 6 (sin (x2 − 3yx)) x2 − 9 (sin (x2 − 3yx)) yx − 3 cos (x2 − 3yx) ∂y∂x ∂ ∂ (i) p(x, y) = 2xy 2 e2yx + 2x2 y 3 e2yx p(x, y) = : 2x2 ye2yx + 2x3 y 2 e2yx ∂x ∂y ∂2 p(x, y) = 2y 2 e2yx + 8xy 3 e2yx + 4x2 y 4 e2yx ∂x2 ∂2 p(x, y) = 2x2 e2yx + 8x3 ye2yx + 4x4 y 2 e2yx ∂y 2 ∂2 p(x, y) = 4xye2yx + 10x2 y 2 e2yx + 4x3 y 3 e2yx ∂y∂x ∂ ∂ 2 2 (j) q(x, y) = −y 2 e−xy + 4x3 y 3 q(x, y) = −2yxe−xy + 3x4 y 2 ∂x ∂y 2 ∂2 ∂ 2 2 4 −xy 2 2 3 q(x, y) = y e + 12x y q(x, y) = −2xe−xy + 4y 2 x2 e−xy + 6x4 y 2 2 ∂x ∂y 2 ∂ 2 2 q(x, y) = −2ye−xy + 2y 3 xe−xy + 12x3 y 2 ∂y∂x (g)