Implicit Partial Differentiation

jdstokes

If there is such a thing. I need to find $\partial z / \partial x$ given $x + y + z = \cosh xyz$. I've never seen the likes of this before and I haven't a clue where to start. Would a reasonable start be to take $\partial /\partial x$ of both sides? If so, it seems like I'm going to end up with an expression in terms of partial differntials.

Related Introductory Physics Homework Help News on Phys.org

OlderDan

Homework Helper
jdstokes said:
If there is such a thing. I need to find $\partial z / \partial x$ given $x + y + z = \cosh xyz$. I've never seen the likes of this before and I haven't a clue where to start. Would a reasonable start be to take $\partial /\partial x$ of both sides? If so, it seems like I'm going to end up with an expression in terms of partial differntials.
The symbol $$\partial z / \partial x$$ is incomplete. We do it all the time, but a symbol for a partial derivative is never complete without specifying what is being held constant when the derivative is taken. You need to decide from the context of your problem what is being held constant here. If z is being held constant, then the partial derivative is zero. On the other hand, if z = z(x,y) and only y is being held constant it is a completely different problem.

jdstokes

All I am given is that z is given implicitly as a function of x and y by the above equation. I take this to mean that y is held constant while x and z are allowed to vary.

saltydog

Homework Helper

$$z^2=Sin[xz]+xy$$

Assuming z is a differentiable function of x and y and I want the partial of z with respect to x, lets take that partial of both sides:

$$\frac{\partial}{\partial x}\{z^2=Sin[xz]+xy\}$$

$$2z\frac{\partial z}{\partial x}=Cos[xz](x\frac{\partial z}{\partial x}+z)+y$$

Isolating the partial:

$$\frac{\partial z}{\partial x}=\frac{zCos(xz)+y}{2z-xCos[xz]}$$

Now, I bet you can do yours!

jdstokes

Thanks for the worked example. I got $\frac{\partial z}{\partial x} = \frac{yz\sinh xyz -1}{1 - xy\sinh xyz}$.

saltydog

Homework Helper
jdstokes said:
Thanks for the worked example. I got $\frac{\partial z}{\partial x} = \frac{yz\sinh xyz -1}{1 - xy\sinh xyz}$.
Put parenthesis around xyz. It's ambiguous as written. Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving