Application of the Implicit function theorem on the function $f(x,y,z)=(x^2+y^2)e^z+\sin(\pi x)yz+2z-1$

Question

I wanted to know if my procedure about this Implicit function theorem question is correct and if my results are correct. Note : I am an engineering student, so I'm mostly concerned about the "computing aspect" of my answer, i.e I just followed the same procedure as in the example our professor gave us. The problem is that the following question is from an old exam from another professor who gives no solution to its exams, so I don't know if my results are correct.

I have the following function : $$f(x,y,z)=(x^2+y^2)e^z+\sin(\pi x)yz+2z-1$$

• (a) Show that there exists a smooth function $g: U \to \mathbb{R}$, which is defined on an open neighbourhood $U \subset \mathbb{R^2} $ of $(1,1)$ and for which $g(1,1)=0$, $f(x,y,g(x,y))=1$ $\forall (x,y) \in U$

• (b) Calculate the derivative $d_{x,y}g(1,1)$
  

---

a) So first I took the partial derivatives of $f$ and got: $$\frac{\mathrm df}{\mathrm dx}= 2xe^z+\pi \cos(\pi x)yz$$ $$\frac{\mathrm df}{\mathrm dy}=2ye^z+\sin(\pi x)z$$ $$\frac{\mathrm df}{\mathrm dz}=e^z(x^2+y^2)+\sin(\pi x)y+2$$ Now, $f(1,1,0)=2-1=1$ and $f_x(1,1,0)=2, f_y(1,1,0)=2$ and $f_z(1,1,0)=4$

Now, given that $f(1,1,0)=1$ and all partial derivatives at $(1,1,0)$ don't yield $0$ (i.e $\nabla f$ at $(1,1,0)$ $\neq$ $(0,0,0)$), there exists a function $g$ that satisfies $f(x,y,g(x,y))=1$ according to the implicit function theorem.
b)
According to the chain rule : $$0 = d_{x,y,z}f(x,y,g(x,y))=d_{x,y}f(x,y,g(x,y)) +d_{z}f(x,y,g(x,y))*d_{x,y}g(x,y)$$ $$-d_{x,y}f(x,y,g(x,y))=d_{z}f(x,y,g(x,y))*d_{x,y}g(x,y)$$ $$-d_{z}f(x,y,g(x,y))^{-1}d_{x,y}f(x,y,g(x,y))=d_{x,y}g(x,y)$$

$$d_{x,y}g(1,1)=-(d_z f(1,1,0)\:)^{-1} \:d_{x,y}f(1,1,0)= - 4^{-1}(2 \: \:\:2)=-\frac{1}{4}(2 \: \:\: 2)$$

Thanks for your help !

Answer 1

Okay, so as said in the comments, everything seems to be correct. I just maybe need to detail a bit more what I did for part b), especially concerning the chain rule, and to be a bit more careful with the notation, if I get a similar question in the future. Thanks for your help.

Edit : as was pointed out in the comments, I just need to take the derivative of the implicit function (i.e. $f_z$), not necessarily the gradient, and check that it's not zero. Also, the derivative notation in part b), $f_{x,y}$, is confusing because one one side, it's the total derivative, while on the other, it's the partial derivative. I could maybe write $f_{x,y,z}$ on the left hand side to make clear that it's the total derivative and keep $f_{x,y}$ on the right hand side to indicate the partial derivatives with respect to x and y.