How to resolve this second partial derivative?: y ∂^3z/(∂u∂v∂w) from z=u(sqrt(v-w))

Question

#y (∂^3z)/(∂u∂v∂w)->z=usqrt(v-w)#

3498 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-01T03:10:32

#=>(∂^3z)/(∂u∂v∂w) = color(blue)(1/(4 (v - w)^(3/2)))#

The derivative are worked from right-to-left as listed in the denominator.

So first, we take the derivative with respect to #w#.

#{partial z} / {partial w} = partial/(partial w){u sqrt(v-w) }#

#= u partial/(partial w){(v-w)^(1/2) }#

# = u(1/2)(v-w)^(-1/2)(-1)#

# = color(orange)(-u/(2 sqrt[v - w]))#

Now we use this result and take its derivative with respect to #v#.

#{partial} / {partial v}{color(orange)(-u/(2 sqrt[v - w]))}= -u/2 {partial} / {partial v}{(v-w)^(-1/2)}#

#= -u/2(-1/2)(v-w)^(-3/2)(1)#

#= color(red)(u/(4 (v - w)^(3/2)))#

Now we use this result and take its derivative with respect to #u#.

# {partial} / {partial u} {color(red)(u/(4 (v - w)^(3/2)))} = 1/(4 (v - w)^(3/2)) {partial} / {partial u} {u}#

# =1/(4 (v - w)^(3/2)) (1) #

# =color(blue)(1/(4 (v - w)^(3/2)) )#

Hence,

#=>(∂^3z)/(∂u∂v∂w) = color(blue)(1/(4 (v - w)^(3/2)))#