Directional Derivative of a Multivariable Function

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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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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