Consider the following. z = x4 + x2y, x = s + 2t − u, y = stu2 Find the following partial derivatives. ∂z ∂s = ∂z ∂t = ∂z ∂u = Find ∂z ∂s , ∂z ∂t , ∂z ∂u when s = 5, t = 1, u = 4. ∂z ∂s = ∂z ∂t = ∂z ∂u =

Given: $z=x^4+x^{2}y, x=s+2t-u, y=st{u}^2$

To find: 
(a) $\frac{dz}{ds}$
(b) $\frac{dz}{dt}$
(c) $\frac{dz}{du}$
(d) $\frac{dz}{ds},\frac{dz}{dt},\text{and}\frac{dz}{du}$ at $s=5,t=1,u=4$

Solution:

We begin by finding the partial derivatives of z with respect to x, y, s, t, and u.

Partial derivative of z with respect to x:
$\frac{dz}{dx}=4x^3+2xy$

Partial derivative of z with respect to y:
$\frac{dz}{dy}=x^2$

Partial derivative of x with respect to s:
$\frac{dx}{ds}=1$

Partial derivative of x with respect to t:
$\frac{dx}{dt}=2$

Partial derivative of x with respect to u:
$\frac{dx}{du}=-1$

Partial derivative of y with respect to s:
$\frac{dy}{ds}=0$

Partial derivative of y with respect to t:
$\frac{dy}{dt}=u^{2}s$

Partial derivative of y with respect to u:
$\frac{dy}{du}=2stu$

Using the chain rule, we can now find the partial derivatives of z with respect to s, t, and u.

Partial derivative of z with respect to s:
$\frac{dz}{ds}=\frac{dz}{dx}\cdot \frac{dx}{ds}+\frac{dz}{dy}\cdot \frac{dy}{ds}$
      $=(4x^3+2xy)\cdot 1+x^2\cdot 0$
      $=4x^3+2xy$
      $=4{(s+2t-u)}^3+2(s+2t-u)\left(st{u}^2\right)$

Substituting $s=5,t=1,\text{and}u=4,$ we get:
$\frac{dz}{ds}=4{(5+2\left(1\right)-4)}^3+2(5+2\left(1\right)-4)(1\cdot 4\cdot 4^2)$
      $=2380$

Partial derivative of z with respect to t:
$\frac{dz}{ds}=\frac{dz}{dx}\cdot \frac{dx}{ds}+\frac{dz}{dy}\cdot \frac{dy}{ds}$
      $=(4x^3+2xy)\cdot 2+x^2\cdot u^{2}s$
      $=8x^3+4xy+x^2{u}^{2}s$
      $=8{(s+2t-u)}^3+4(s+2t-u)\left(st{u}^2\right)+{(s+2t-u)}^2(s\cdot t\cdot 4^2)$

Substituting $s=5,t=1,\text{and}u=4$, we get:
$\frac{dz}{dt}=8{(5+2\left(1\right)-4)}^3+4(5+2\left(1\right)-4)(1\cdot 4\cdot 4^2)+{(5+2\left(1\right)-4)}^2(5\cdot 1\cdot 4^2)$
      $=3388$

Partial derivative of z with respect to u:
$\frac{dz}{du}=\frac{dz}{dx}\cdot \frac{dx}{du}+\frac{dz}{dy}\cdot \frac{dy}{du}$
      $=(4x^3+2xy)\cdot (-1)+x^2\cdot 2stu$
      $=-4x^3-2xy+2x^{2}stu$
      $=-4{(s+2t-u)}^3-2(s+2t-u)\left(st{u}^2\right)+2{(s+2t-u)}^2\left(stu\right)\cdot 4$

Substituting $s=5,t=1,\text{and}u=4,$ we get:
$\frac{dz}{du}=-4{(5+2\left(1\right)-4)}^3-2(5+2\left(1\right)-4)(1\cdot 4\cdot 4^2)+2{(5+2\left(1\right)-4)}^2(1\cdot 4\cdot 4)\cdot 4$
      $=-812$

Therefore, the partial derivatives of z are:
(a) $\frac{dz}{ds}=4{(s+2t-u)}^3+2(s+2t-u)\left(st{u}^2\right)=2380$
(b) $\frac{dz}{dt}=8{(s+2t-u)}^3+4(s+2t-u)\left(st{u}^2\right)+{(s+2t-u)}^2(s\cdot t\cdot 4^2)=3388$
(c) $\frac{dz}{du}=-4{(s+2t-u)}^3-2(s+2t-u)\left(st{u}^2\right)+2{(s+2t-u)}^2\left(stu\right)\cdot 4=-812$

Hence, the partial derivatives of z with respect to s, t, and u at $s=5,t=1,\text{and}u=4$ are $\frac{dz}{ds}=2380,\frac{dz}{dt}=3388,\text{and}\frac{dz}{du}=-812,$ respectively.