Two variable calculus - finding partial derivative = 0

[Njords]

I'm guessing we ignore x=1 since not in the domain from part a)

so I'm stuck with part c) assuming I've found the correct partial derivative.

Only solutions for real numbers since that's all that's covered in the scope of the lectures.

Other similar problems are set out solving as simultaneous equations but they were way easier than this....

 

[HallsofIvy]

Assuming x and y are to be real numbers, the domain is x> 1. You say you are "stuck on c assuming I've found the right partial derivatives". But (b) asks for all second derivatives. You only show the first derivatives. Do you mean that you have found the second derivatives by are only showing the first derivatives because you want to determine where they are 0?

You have \(\displaystyle f_x= -2(x- 1)^{-2}e^{y^3}+ \frac{1}{2}(x- 1)^{-1/2}\) and \(\displaystyle f_y= 6y^2e^{y^3}(x- 1)^{-1}\). It should be clear that the second expression is 0 only for y= 0: \(\displaystyle e^{y^3}\) and \(\displaystyle (x- 1)^{-1}\) are never 0. Since y must be 0 in order that the second expression be 0, to find x, y such that both expressions are 0, we can set y= 0 in the first expression: \(\displaystyle -2(x- 1)^{-2}+ \frac{1}{2}(x- 1)^{-1/2}= 0\). Then \(\displaystyle -2(x- 1)^{-2}= -\frac{1}{2}(x- 1)^{-1/2}\). Get rid of the square root by squaring both sides (remembering that this may introduce "spurious solution": \(\displaystyle 4(x- 1)^{-4}= \frac{1}{4}(x- 1)\). Multiply both sides by \(\displaystyle 4(x- 1)^4\): \(\displaystyle (x- 1)^5= 16\).

 

[Njords]

I did notice the second one y=0 would make the whole thing = 0 but second guessed myself when using that in first expression.

so then x= 16^(1/5)+1 , y=0 is the answer ...

thanks