# Find the equation to the tangent plane

$f(x,y) = \sqrt{xy}$ 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

• Er...what you got and what is in your book is exactly the same after you multiply by two your equation... – DonAntonio Jun 14 '16 at 17:55
• Yep. Apply $-z$, then $\cdot 2$ and you get your result. – Maximilian Gerhardt Jun 14 '16 at 17:56
• The two equations of the plane are equivalent. – Doug M Jun 14 '16 at 17:56

$$\frac{1}{2}(x-1)+\frac{1}{2}(y-1)=z-1$$ $$z*2-1*2=\frac{1}{2}(x-1)*2+\frac{1}{2}(y-1)*2$$ $$2z-2=y+x-2$$ $$2z=y+x$$ $$x+y-2z=0$$