The system of equations is given:

$$\left\{ \begin{array}{c} x=u+\ln(v) \\ y=v-\ln(u) \\ z=2u+v \end{array} \right. $$

I need to find the partial derivative of z with respect to x at u=1 and v=1 What I did is as follows: I applied total differentiation and then divided the entire system by ∂x and got the answer=3, but it does not overlap with the correct answer.

Edited: The options for correct answers are given i9n this pic: pic

  • $\begingroup$ Is your way the following? $\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}=2\cdot\frac{1}{1}+1\cdot1=3.$ $\endgroup$ – Michael Rozenberg Jul 8 '19 at 10:44
  • $\begingroup$ @MichaelRozenberg yep similar $\endgroup$ – Mina Babayeva Jul 8 '19 at 10:56
  • $\begingroup$ Is the correct answer 1.5? $\endgroup$ – johnnyb Jul 10 '19 at 12:47
  • $\begingroup$ I could not define it because of the wrong answer choice, please see the pic added to the edited post $\endgroup$ – Mina Babayeva Jul 10 '19 at 13:28

The answer is $\frac{\partial z}{\partial x} = \frac{3}{2}$ and $\frac{\partial z}{\partial y} = -\frac{1}{2}$.

So, let's start with the initial set of equations: $$ x = u + \ln(v) \\ y = v - \ln(u) \\ z = 2u + v $$ Now, it is unspecified, but I'm going to presume, based on the variable names chosen, that $x$, $y$, and $z$ are independent variables, and $u$ and $v$ are dependent variables. That \emph{really} matters, \emph{especially} with partial derivatives.

Now, to begin with we are going to take the \emph{total} differential of all of these: $$ dx = du + \frac{dv}{v} \\ dy = dv - \frac{du}{u} \\ dz = 2du + dv $$

Now, the question stipulates that we are finding these things where $u = 1$ and $v = 1$, so we can substitute them at this time (we couldn't substitute them before, because we needed to know how everything is varying with respect to each other, but now that we know this, we can now use these variables directly). So, our equations become

$$ dx = du + dv \\ dy = dv - du \\ dz = 2du + dv $$

We want to solve for $\frac{\partial z}{\partial x}$. However, the \emph{dependent} variables are still allowed to change, because they could be influenced by $x$ or $z$. Essentially, we are just holding $y$ still, but the others are free. I'm going to use $\partial$ notation, but just note that $\partial$ notation can be misleading (see here). If $y$ is still, that means that $dy = 0$. Thus we get

$$ \partial x = \partial u + \partial v \\ 0 = \partial v - \partial u \\ \partial z = 2 \partial u + \partial v $$

We can simplify the second line to say $\partial v = \partial u$. This allows the following simplification:

$$ \frac{\partial x}{2} = \partial u \\ \partial z = 3\partial u \\ \therefore \frac{\partial z}{\partial x} = \frac{3}{2} $$

We can do a similar procedure for $\frac{\partial z}{\partial y}$ and will find the answer is $-\frac{1}{2}$:

$$ dx = du + dv \\ dy = dv - du \\ dz = 2du + dv \\ ~\\ 0 = \partial u + \partial v \\ \partial y = \partial v - \partial u \\ \partial z = 2\partial u + \partial v \\ ~\\ \partial u = - \partial v \\ \partial y = 2\partial v \\ \partial z = -\partial v \\ \partial z = -\frac{\partial y}{2} \\ \frac{\partial z}{\partial y} = -\frac{1}{2} $$

Just to note, the partials in this second series of equations \emph{cannot} be intermixed with the partials in the first series. See the link above for why this is the case.

If this answer was helpful, do you mind letting me know where the question is from?

  • $\begingroup$ I wondered what the answer would be for mixed partial derivative z with respect to x and y, dz/(dxdy), but I could not figure it out, can You suggest something useful? $\endgroup$ – Mina Babayeva Jul 15 '19 at 4:59

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