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First partial derivatives.

Let f be a function of two variables x,y, defined on a neighborhood of the point P0=(x0,y0).

Definition 5.1       

For a function of more than two variables, we define the other partail derivatives by the same way.

In each case, we consider one variable as a constant and differentiate with respect to the other variable, according to the well-known rules of differentiation, as learnt in Calculus I.

Example 5.2   Let $f(x)=x^2y^3+\cos xy$. Then at any point $P \in \mathbb{R} ^2$, we have:

\begin{displaymath}\frac {\partial f}{\partial x}\Bigg\vert_{P} = 2xy^3 - y \sin...
... {\partial f}{\partial x}\Bigg\vert_{P} = 3x^2y^2 - x \sin xy.
\end{displaymath}

Example 5.3   Let $f(x)=\frac {x-5y}{x^2+y^2}$. Then at any point $P \in \mathbb{R} ^2- \{ (0,0) \}$, we have:
\begin{align*}\frac {\partial f}{\partial x}\Bigg\vert_{P} &= \frac {1 \; ( x^2+...
...- (x-5y)(2y)}{(x^2+y^2)^2}
= - \frac {5x^2+2xy-5y^2}{(x^2+y^2)^2}.
\end{align*}

Example 5.4   Find $\frac {\partial z}{\partial x}$ when z is defined as an implicit function of x and y by the relation $xz + \ln z = 2x +3y$.

We differentiate both sides of the relation with respect to x:
\begin{align*}\frac {\partial }{\partial x} (xz + \ln z ) & = \frac {\partial }{...
...} \qquad \frac {\partial z}{\partial x} & = \frac {2z-z^2}{zx + 1}.
\end{align*}

Notations: $f_x = \frac {\partial f}{\partial x}$ and $f_y = \frac {\partial f}{\partial y}$.


next up previous contents
Next: Differentiability, linearization. Up: Partial derivatives; the differential Previous: Partial derivatives; the differential
Noah Dana-Picard
2001-05-30