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$f(x,y) = \sqrt{xy}$ at the point (1,1,1)

$f_x$(1,1) = $\frac{\sqrt{y}}{2\sqrt{x}}$ = $\frac{1}{2}$

same for $f_y$

setting up the formula I get:

$\frac{1}{2}$$(x-1)$+$\frac{1}{2}$$(y-1)$$=$$z-1$ and simplifies to $\frac{1}{2}x$+$\frac{1}{2}y$ $=$ $z$

I must be doing something wrong in my setup as the book says $x+y-2z=0$ is the equation

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    $\begingroup$ Er...what you got and what is in your book is exactly the same after you multiply by two your equation... $\endgroup$ – DonAntonio Jun 14 '16 at 17:55
  • $\begingroup$ Yep. Apply $-z$, then $\cdot 2$ and you get your result. $\endgroup$ – Maximilian Gerhardt Jun 14 '16 at 17:56
  • $\begingroup$ The two equations of the plane are equivalent. $\endgroup$ – Doug M Jun 14 '16 at 17:56
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$$\frac{1}{2}(x-1)+\frac{1}{2}(y-1)=z-1$$ $$z*2-1*2=\frac{1}{2}(x-1)*2+\frac{1}{2}(y-1)*2$$ $$2z-2=y+x-2$$ $$2z=y+x$$ $$x+y-2z=0$$

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  • $\begingroup$ Yes I realized that after I had posted this. Thank you. $\endgroup$ – K. Gibson Jun 16 '16 at 19:34

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