# Partial Derivatives Homework

I just started learning partial derivatives on my own, and I get confused about concepts such as independence and dependence of variables.

For example I needed to find du/dx given the following equation: $u = x^u + u^y$

What I did was to rearrange the equation to: $\ln u / u = \ln x / (1-y)$

Then I expressed the partial derivative as: $du/dx = u^2 / (x(1-y)(1-\ln u))$

I’m not sure how to find the derivative implicitly without separating u (the dependent variable) from x, y (the independent variables).

The reason I rearranged the equation was because I thought that one couldn’t treat the ‘u’ terms on the right side of the equation as constants (during the partial differentiation) since they are dependent on x.

Clarifications would be greatly appreciated!

• I'm dubious of that first step, the rearrangement. See what happens when $u=2$, $x=1$, $y=0$. – Gerry Myerson Mar 2 '13 at 5:37
• Sorry, you're right I made a stupid mistake. All the same, how should I approach this. I can get x in terms of u and y, or y in terms of u and x. But, I'm not sure if I can treat y or x as constants in both cases respectively, since they can be expressed in terms of each other. – user61209 Mar 2 '13 at 9:09
• The whole point in implicit differentiation is to differentiate even though it's not possible to express a function only in terms of its variables. Maybe it'll help if you take a look at some examples, like #3 here: en.wikipedia.org/wiki/Implicit_differentiation#Examples_2 – Wheepy Mar 2 '13 at 15:56

Look at it this way: You have three variables $u$, $x$, $y$ (modeling, e.g., physical quantities) that are related by the "constituent equation" $$\bigl(F(x,y,u):=\bigr)\qquad x^u+u^y-u=0\ .\tag{1}$$ An admissible state is, e.g., the point ${\bf p}_0:=(1,0,2)$. The equation $(1)$ reduces the number of degrees of freedom from $3$ to $2$; therefore we may nourish the idea that given $x$ and $y$ the value of $u$ is determined. In other words: There is an underlying function $$\phi:\quad(x,y)\mapsto u=\phi(x,y)\ ,$$ which produces for all suitable $(x,y)$ an $u$ such that the triple ${\bf p}:=\bigl(x,y,u\bigr)$ is admissible. This means that $$F\bigl(x,y,\phi(x,y)\bigr)=0\quad\forall x,\ \forall y\ .$$ Taking partial derivatives with respect to $x$ and $y$, and using the chain rule we can deduce that $$F_x\bigl(x,y,\phi(x,y)\bigr)+F_u\bigl(x,y,\phi(x,y)\bigr)\phi_x(x,y)\equiv0,\quad F_y\bigl(x,y,\phi(x,y)\bigr)+F_u\bigl(x,y,\phi(x,y)\bigr)\phi_y(x,y)\equiv0\ .$$ Solving for the partial derivatives of $\phi$ we get $$\phi_x(x,y)=-{F_x\bigl(x,y,\phi(x,y)\bigr)\over F_u\bigl(x,y,\phi(x,y)\bigr)}, \quad \phi_y(x,y)=-{F_y\bigl(x,y,\phi(x,y)\bigr)\over F_u\bigl(x,y,\phi(x,y)\bigr)}\ .$$ This can be interpreted as follows: Near any admissible point ${\bf p}_0=(x_0,y_0,u_0)$ the "local" function $(x,y)\mapsto u:=\phi(x,y)$ has partial derivatives $${\partial u\over\partial x}=-{F_x({\bf p})\over F_u({\bf p})}, \quad {\partial u\over\partial y}=-{F_y({\bf p})\over F_u({\bf p})}\ ,\tag{2}$$ under the essential condition that $F_u({\bf p}_0)\ne0$.
In our example $F_u(x,y,u)=\log x\cdot x^u- y u^{y-1}-1$; so at the point ${\bf p}_0:=(1,0,2)$ we have $F_u({\bf p}_0)=-1$. Therefore the formula $(2)$ is applicable at ${\bf p}_0$. I leave it to the OP to compute the values of the partial derivatives appearing therein.
For this problem you need to use implicit differentiation. So you treat $u$ as a function of the variable $x$ and you treat $y$ as a constant. So you could write $u = f(x)$ if that makes it easier. So you find $$\frac{d}{dx} f(x) = \frac{d}{dx} x^{f(x)} + f(x)^y \quad \Rightarrow \\ f'(x) = f'(x)\ln(x) + \frac{f(x)}{x} + \ln(y)f(x)^yf'(x).$$ Now you solve this for $f'(x) = u'$.