Partial derivatives

Whatupdoc · Sep 30, 2005

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?

Galileo · Oct 1, 2005

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.

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Partial derivatives

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