Partial derivatives

  • Thread starter Whatupdoc
  • Start date
  • #1
99
0
If sin (2x+4y+z) = 0 , find the first partial derivatives [tex]\frac{dz}{dx}[/tex] at the point (0,0,0)

A.) [tex]\frac{dz}{dx}(0,0,0) =[/tex] _________________

isnt this saying get the derivative of z, respect to x? i'm just kinda confuse since the variable 'z' is also in the problem.


well i got the derivative of that function with respect to x and got 2*cos(2x+4y+z), plugged in 0,0,0 and got 2*cos(0), which is wrong. the answer should be -2*cos(0), where did the negative sign come from?
 

Answers and Replies

  • #2
Galileo
Science Advisor
Homework Helper
1,989
6
Judging from the question, You should regard z as a function of x (and possibly y).
So you are given that:

[tex]\sin(2x+4y+z(x,y))=0[/tex]

Now use partial differentiation to find dz/dx.
 

Related Threads on Partial derivatives

  • Last Post
Replies
5
Views
8K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
16
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
10K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
979
Top