# Implicit partial derivative

1. Feb 10, 2015

### Calpalned

1. The problem statement, all variables and given/known data
Find (∂z/∂x) of 6xyz

2. Relevant equations
N/a

3. The attempt at a solution
The correct answer is 6xy(∂z/∂x) but I would like proof of it. I got something different when I tried taking the partial derivative.

6xyz = 6x(yz) = Multiplication rule for derivatives

6(∂x/∂x) + y(∂z/∂x)

What did I do wrong? Thanks

2. Feb 10, 2015

### Staff: Mentor

Is this the complete problem statement? Generally speaking you don't find "∂z/∂x" of something, and more than you find "dy/dx" of something. You can, however, find the derivative with respect to, say, x of a function (denoted d/dx(f)) or the partial of some function with respect to, say z (denoted ∂/∂x(f).

The symbols ∂z/∂x and dy/dx represent derivatives, whereas the symbols ∂/∂x and d/dx represent operators that can be applied to functions.

Please verify that what you're providing is the complete problem statement.

3. Feb 11, 2015

### Calpalned

The full question is find (∂z/∂x) of x3 + y3 + z3 + 6xyz = 1
My textbook says the partial derivative is 3x2 + 3z2(∂z/∂x) + 6xy(∂z/∂x) = 0
I don't get how to take the derivative of the red part.

4. Feb 11, 2015

### Calpalned

The derivative above is from my textbook's solutions guide.

5. Feb 11, 2015

### Staff: Mentor

Corrected to: find (∂z/∂x) of if x3 + y3 + z3 + 6xyz = 1

It looks to me like there's a mistake in the solution guide.
$∂/∂x(x^3) = 3x^2$ and $∂/∂x(y^3) = 0$ and $∂/∂x(z^3) = 3z^2 ∂z/∂x$

BUT
$∂/∂x(6xyz) = 6yz$ if z is independent of x, in which case ∂z/∂x would be 0. However, if z is dependent on x, you need to use the product rule. Since you're asked to find ∂z/∂x, it must be the case that z is a function of x (is dependent on x).