Finding the partial differential of given problem

Question

Given:

$$u=ax+bx^2+u^4$$

Find the partial derivative with respect to $x$ and $y$ and as well total derivative of $u$.

Please tell me complete description of this question.

Answer 1

In answering this question, I am going to assume at the beginning that $u = u(x, y)$ is a function of $x$ and $y$, and that it is sufficiently smooth that the differentiations I will carry out make sense. Probably it will suffice to take $u \in C^1(D, \mathbb {R})$, where $D \subset R^2$ is an open region in the $x-y$ plane $R^2$. After deriving some basic formulas for various derivatives of $u$, I will go back and discuss the differentiability conditions on $u$ in more detail.

Throughout, I will use subscript notation to denote partial derivatives, thus we have $u_x = \frac{\partial u}{\partial x}$, $u_y = \frac{\partial u}{\partial y}$ $u_{xy} = \frac{\partial u}{\partial x \partial y}$ and so forth.

The key thing to remember when taking partial derivatives with respect to a given variable, say $x$, is that we hold all other independent variables, i. e. $y$ in this case, constant; given that assumption, the ordinary rules of single -variable calculus apply. Now in the present situation we have

$u = ax + bx^2 + u^4$,

so that a $u_x = a + 2bx + 4u^3u_x$,

or

$(1 - 4u^3)u_x = a + 2bx$,

which yields

$u_x = \frac{a + 2bx}{(1 - 4u^3)}$,

which is a valid expression for $u_x$ as long as $1 - 4u^3 \ne 0$; if we knew the value of $u(x, y)$ at some point $(x, y) \in D$, we could calculate the value of $u_x$ there, even without knowing the exact functional form of $u$ in terms of $x$ and $y$, provided we were given the value of $u$, and that, of course, $1 - 4u^3 \ne 0$. A similar computation yields $u_y$: differentiating the given equation with respect to $y$ it follows that

$u_y = 4u^3u_y$,

or

$(1- 4u^3)u_y = 0$;

apparently $u_y = 0$ as long as $4u^3 \ne 1$; but since we have to stay away from points where $4u^3 = 1$ in any event, to obtain $u_x$, this condition on $u$ does not seem too onerous or overly restrictive. To summarize , then, we can say that as long as $4u^3 - 1 \ne 0$, we have, for the partial derivatives of $u$:

$u_x = \frac{a + 2bx}{(1 - 4u^3)}$,

$u_y = 0$;

these expressions combine to allow us to write the gradient of $u$, $\nabla u$, as

$\nabla u = (u_x, u_y) = (\frac{a + 2bx}{(1 - 4u^3)}, 0)$.

As for the total derivative of $u$, this is a concept with many possible interpretations, see for example the wikipedia entry http://en.m.wikipedia.org/wiki/Total_derivative. For example, the total derivative of $u$ along a differentiable curve $(x(t), y(t))$ which avoids points such that $1-4u^3 =0$ can be taken to be

$\frac{du(x(t), y(t))}{dt} = \nabla u \cdot (\dot{x}(t), \dot{y}(t)) = \frac{a + 2bx(t)}{(1 - 4u^3(x(t), y(t))}\dot{x}(t)$.

It should perhaps come as no surprise that $u_y = 0$, since the equation we started with,

$u = ax + bx^2 + u^4$,

essentially defines $u$ as a function of $x$, in the sense that $u(x)$ must be a root of this quartic equation the degree $0$ term of which is itself parametrized by the variable $x$.