Partial Derivative Calculations for 2xy + 4yz + 5xz with Chain Rule

[olivia333]

Homework Statement

w = 2xy + 4yz + 5xz
x = st
y = 3^(st)
z = t^2

s=5
t=1

Homework Equations

Chain rule: xy = x*y' + y*x'

The Attempt at a Solution

w = 2ste^st + 4te^st + 5st³

(partial derivatives) dw/dt = 2s²te^st + 2se^st + 4ts^st + 4e^st + 15st²

(partial derivatives) dw/dt (5,1) = 2(5)²e⁵ + 2*5e⁵ + 20e⁵ + 4e⁵ + 75

= 84e⁵+75

This is not correct. What did I mess up on? Thanks!

 

[HmBe]

When you said, y = 3^(st), did you mean y = e^(st)?

 

[olivia333]

Yes I did mean that, but I actually figured out what I did wrong. I subbed in a y as t and not t^2.

 

[HmBe]

Your 4yz term is not correct.

EDIT: Yep, beat me to it.