Finding Partial Derviatives
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Uploader Comments (patrickJMT)
All Comments (183)
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Thank you, you made the concept very understandable.
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Damn. i need a lecturer like you. mine sucks.
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When you took the derivative of x*y^2*z^3 you grouped (x)(y^2*z^2)...so why didnt you treat it with the product rule ?
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@f3rstl13 the first term times the derivative of the second + the derivative of the first times the second. in the "first" we leave the r alone and multiply by the derivative of the second (where now r is indeed a constant) the we would have to add the derivative of the first times the second but if this time we actually take the derivative of r knowing that it is a constant , its derivative is ZERO. so it would cancel out the whole second part which i think is how it is.But I may be wrong.
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it's making more sense to me know. thank you
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Also, if the chain rule is u*dv + v*du, how does that work when you've got 3 functions of r? You've got r, r^2 and ln(r^2). Don't I need more than u and v? Don't I need u, v, and w? I was able to solve when I used v = w*x, w = ln(x) and x = r^2 + s^2 then dv = dw*dx
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When you're solving the f(r,s) function for partial s (aorund the 5:59 mark) I don't understand how the first "r" is kept as a constant but the second r (r^2) is zero. Is the derivative of a constant a constant when you're multiplying but zero when you're adding?
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CHEERS!
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Your handwriting is so nice.
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Right now I'm studying to become a high school math teacher. Not only am I learning calculus by watching your videos, but I am also learning how to be an effective teacher. You are building futures, Patrick.
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I dont understand this at all! How does making it 0 make any sense?
zarp29 2 months ago
@zarp29 what is ' it ' ?
patrickJMT 2 months ago
you're my first man crush
green681 5 months ago 21
@green681 awwww baby
patrickJMT 4 months ago 35
thanks man
emtcrystal 5 months ago
@emtcrystal no problem!
patrickJMT 5 months ago