Question

Use the chain rule to find the indicated partial derivatives. z = x4 + x2y, x = s + 2t − u, y = stu2; ∂z ∂s , ∂z ∂t , ∂z ∂u when s = 3, t = 1, u = 2

Answer 1

By the chain rule,

$\dfrac{\partial z}{\partial s}=\dfrac{\partial z}{\partial x}\cdot\dfrac{\partial x}{\partial s}+\dfrac{\partial z}{\partial y}\cdot\dfrac{\partial y}{\partial s}$

$\dfrac{\partial z}{\partial s}=(4x^3+2xy)(1)+(x^2)(tu^2)$

When $(s,t,u)=(3,1,2)$, we get $(x,y)=(3,12)$, so at this point the partial derivative of $z$ with respect to $s$ takes on a value of

$\dfrac{\partial z}{\partial s}\bigg|_{s=3,t=1,u=2}=216$

Answer 2

Using the chain rule, the derivatives are:

$\frac{\partial z}{\partial s} = 216$

$\frac{\partial z}{\partial t} = 576$

$\frac{\partial z}{\partial u} = -288$

The function z is:

$z = x^4 + x^2y$

It is a function of x and y, which are functions of s, t and u. Thus, the partial derivatives are:

$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$

$\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}$

$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$

The functions for x and y are:

$x = s + 2t$

$y = stu^2$

Then:

$\frac{\partial z}{\partial s} = (4x^3+2xy) + x^2(tu^2)$

$\frac{\partial z}{\partial s} = (4(s + 2t - u)^3+2(s + 2t - u)(stu^2)) + (s+2t - u)^2(tu^2)$

Considering s = 3, t = 1, u = 2:

$\frac{\partial z}{\partial s} = (4(3 + 2 - 2)^3+2(3 + 2 - 2)(3(1)(2^2)) + (3+2-2)^2(2^2) = 216$

$\frac{\partial z}{\partial t} = 2(4x^3+2xy) + x^2su^2$

$\frac{\partial z}{\partial t} = 2(4(s + 2t - u)^3+2(s + 2t - u)(stu^2)) + (s+2t - u)^2su^2$

$\frac{\partial z}{\partial t} = 2(4(3 + 2 - 2)^3+2(3 + 2 - 2)(3(1)(2^2)) + (3+2 - 2)^2(3)(2^2) = 576$

$\frac{\partial z}{\partial u} = -(4x^3+2xy) + 2x^2stu$

$\frac{\partial z}{\partial u} = -(4(s + 2t - u)^3+2(s + 2t - u)(stu^2)) + 2(s + 2t - u)^2stu$

$\frac{\partial z}{\partial u} = -(4(3 + 2 - 2)^3+2(3 + 2 - 2)(3(1)(2^2)) + 2(3 + 2 - 2)^2(3)(1)(2) = -288$

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