Question related to a derivative of function defined implicitly

Question

Problem

The equation $x+z+(y+z)^2=6$ defines z implicitly as a function of x and y ,say z=f(x,y) . Compute the partial derivative $\frac{\partial f}{\partial x}$

Attempt

Let $g$ be a auxiliary function defined as follows

$g(x,y)=F(x,y,f(x,y))=0$

Then,

$\frac{\partial g}{\partial x}= \frac{\partial F}{\partial x}+\frac{\partial F}{\partial z}\frac{\partial f}{\partial x}$

and $\frac{\partial g}{\partial y}= \frac{\partial F}{\partial y}+\frac{\partial F}{\partial z}\frac{\partial f}{\partial y}$.

After this my approach is to equate both $\frac{\partial g}{\partial x}$ and $\frac{\partial f}{\partial y}$ to zero but i don't have any logical reason for doing it.

My previous related to this doubt of mine. 1

Please help

Answer 1

Your approach is correct: indeed if you have $F(x,y,f(x,y))=0$ deriving both sides gives you $\partial_xF=\partial_x0=0$ and the same for $y$, and this is allowed because you are "doing the same on both sides".

---

Now the complete approach: Just substitute $z=z(x,y)$ in the equation and bring everything to the left:

$$F(x,y,z(x,y))=x+z(x,y)+(y+z(x,y))^2-6=0$$

Now derive for $x$: this is a linear operation and thus we can derive both sides of the equation. Moreover $y$ is independent so $\partial_x y=0$ and thus $$\partial_x(F(x,y,z(x,y)))=\partial_xF+\partial_zF\cdot\partial_xz=1+2(y+z(x,y))\cdot \partial_xz=0$$ Thus $$\partial_x z=\frac1{1-2(y+z(x,y))}=\frac1{1-2(y+z)}.$$ Now you either solve this PDE or you substitute $z$ via solving the previous equation (which is just a quadratic equation in $z$).

Answer 2

Since you wrote this as "z= f(x,y)", $\frac{\partial f}{\partial x}$ is the same as $\frac{\partial z}{\partial x}$. Now, z is a function of the two independent variables x and y. That means that y is NOT a function of x.

$\frac{\partial}{\partial x}\left(x+ (y+ z)^2\right)= \frac{\partial x}{\partial x}+ \frac{\partial (y+ z)^2}{\partial x}= 1+ 2(y+ z)\frac{\partial y+ z}{\partial x}= 1+ 2(y+ z)\frac{\partial z}{\partial x}$ where I have used, immediately that, since y is independent of x, its derivative with respect to x is 0. The right side of the equation, 6, is a constant so its derivative is 0: $1+ 2(y+ z)\frac{\partial z}{\partial x}= 0$ so $\frac{\partial z}{\partial x}= \frac{\partial f}{\partial x}= \frac{-1}{2(y+ z)}$