\[\text {Find the partial derivatives with respect to z, x, and y.}\] \[2x^2 + 4xy + z^2 = 25\] @Luigi0210 enjoy.
So: \[\LARGE f_{z}:3z^2=0\]?
Find dz/dx with respect to x, treat y as a constant.
that has nothin to do with dis problem
basically you differentiate implicitly
Sorry, my head's still in the gutter
Hmm, I think I looked at the wrong video then ._.
Teach me step-by-step sham.
\[2x^2 + 4xy + x^2 = 25\] dz/dx with respect to x. treat y as a constant. \[4x + 4y + 2zz' = 0\] \[z'(2z) = -4(x+y)\] \[z' = \frac{ -2(x+y) }{ z }\]
make sense?
you don't use the product rule for 4xy because we are treating y as a constant :P
Other than that x^2 in the beginning that's suppose to be a z.. I follow xD
huh?
well anyways, i know you can do implicit diff try it for dy/dx
remember dy/dx means with respect to x. so this time z is a constant :) and the derivative of a constant is zero.
\[\LARGE 2x^2 + 4xy + z^2 = 25\] \[\LARGE 4x+4y'+z^2=0\] Am I right so far..? Probably not..
nope, you treat z as a constant
\[4x + 4y + 4xy' = 0 \] and you forgot the product rule too
Ugh, I'm not focused enough..
\[y' = \frac{ -(x+y) }{ x }\]
What notes/websites did you use to learn this stuff?
not much really. i used my calc book mostly. i looked at this tho http://tutorial.math.lamar.edu/Classes/CalcIII/PartialDerivatives.aspx
What's the name of dat calc book of yours anyways? xD
oh and the notation for y' is \[f_y (x,y,z) = \frac{ -(x+y) }{ x }\] \[f_z (x,y,z) = \frac{ -2(x+y) }{ z }\]
and z*
i'm using thomas transcendental 10th edition and james stewart 5e i think
Alrighty, let me get more familiar with the topics and such and I'll get back to y'all on it~
do the last one! d/dx
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