Directional Derivative of a Multivariable Function

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SUMMARY

The discussion focuses on finding the directional derivative of the multivariable function f(x,y) = y²ln(x) at the point P(1,4) in the direction of the vector a = -3i + 3j. The user initially calculated the normalized direction vector and applied the partial derivatives but encountered discrepancies in their results. After reviewing the calculations, it was confirmed that the correct answer is -8√2, which is equivalent to -16/√2. The user clarified that their confusion stemmed from misinterpreting the simplification of the results.

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Homework Statement


Find the directional derivative of f at P in the direction of a.
f(x,y)=y2lnx; P(1,4); a= -3i + 3j

Homework Equations





The Attempt at a Solution


I first normalized the vector and got an = -3/(18)^1/2 + 3/(18)^1/2
Then i took the partial derivative and multiplied it by the vector
y2/X(-3/181/2) + 2(y)lnx(3/181/2
Ln(1) = 0 so the whole right side is gone, and after i plug in 1 and 4 on the right side i get -16/21/2. The answer should be -8/21/2. I've done 5 problems like this now and each time I'm off by a negative or a factor or something like that. What am i doing wrong?
 
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I get -16/sqrt(2) as well. Are you sure the answer wasn't -8*sqrt(2)? That would be the same thing you have.
 
Thanks, i just went back and checked the back of the book, the answer is supposed to be -8*sqrt(2) not -8/sqrt(2). Also i didn't notice that -16/sqrt(2) is the same thing as -8*sqrt(2). Thanks for the help.
 

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