Calculo 3 derivadas parciales K

a)

wx = 2xy+yz*2+2xz

wy = V^2+xz^2

wz = ×^2

b)

Wx = 2x/[z·ln(×^2+y^2)]

Wy = 2y/z·ln(×^2+y^2)

Wz =-ln(×^2+y^2)/z^2

c)

fx = 2x/z

fy = 2y/z

fz = -(×^2+y^2)/z^2

d)

fx = 2y^z

fy = 2xzy^(z-1)

fz = 2xy^z·ln(y)

e) f =xsen(yz) + ysen(xz)

fx = sen(yz) + yzcos(xz)

fy = xzsen(yz) + sen(xz)

fz = xysen(yz) +xysen(xz)

f)

fx = 2xyz -z

fy = zx^2

fz = yx^2-x

g)

gw = w/sqrt(w^2+t^2+z^2)

gt = t/sqrt(w^2+t^2+z^2)

gz = z/sqrt(w^2+t^2+z^2)

h)

hu = 2u

hv = 2v

hw = -t/wt =-1/w

ht = = -w/wt =-1/t

i)

Tx = [yz·sqrt(×^2+y^2+z^2) - ×^2·yz/sqrt(×^2+y^2+z^2)] /(×^2+y^2+z^2) =

[yz(×^2+y^2+z^2)-x^2·yz]/(×^2+y^2+z^2)^(3/2) =

yz(y^2+z^2)/(×^2+y^2+z^2)^(3/2)

Ty = xz(×^2+z^2)/(×^2+y^2+z^2)^(3/2)

Tz = xy(×^2+y^2)/(×^2+y^2+z^2)^(3/2)

Y eso es todo.