# Find partials for $f$ and the equation of plane

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1. Oct 8, 2016

### toforfiltum

1. The problem statement, all variables and given/known data

Function is $f(x,y)=((x-1)y)^\frac{2}{3}$,$\space\space(a,b)=(1,0)$

a) Calculate $f_{x}(a,b)$ and $f_{y}(a,b)$ at point $(a,b)$ and write the equation for the plane.

2. Relevant equations

3. The attempt at a solution
So,

$f_{x}(x,y)=\frac{2y^\frac{2}{3}}{3(x-1)^\frac{1}{3}}$

$f_{y}(x,y)=\frac{2(x-1)^\frac{2}{3}}{3y^\frac{1}{3}}$

Therefore, at $(1,0)$, $f_{x}(x,y)=0$ and $f_{y}(x,y)=0$

Equation of plane at $(1,0)$ is $z=0$.

What I'm doing seems right, but there is a nagging feeling that I'm wrong.

So thoughts?

Thanks.

Last edited: Oct 8, 2016
2. Oct 8, 2016

### PeroK

Your differentiation of $x^{\frac23}$ is wrong.

3. Oct 8, 2016

### Ray Vickson

Yes, indeed, both of your partial derivatives are incorrect. I don't get $\frac{2}{3} - 1 = 3$, as you seem to have written.

4. Oct 8, 2016

5. Oct 8, 2016

### Staff: Mentor

The corrected partials are now correct, but the values you show at (1, 0) aren't. Notice that in both cases, both the numerator and denominator are zero.

6. Oct 8, 2016

### toforfiltum

Is it undefined in both cases because they involve a division by zero?

7. Oct 8, 2016

### Staff: Mentor

Yes. Both partials are undefined at (1, 0).

8. Oct 8, 2016

### toforfiltum

So that means there is no equation that exists for the tangent plane at that point, right?

9. Oct 8, 2016

### Staff: Mentor

Yes

10. Oct 8, 2016

Thanks!