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Revision of Partial Differentiation

We revise partial differentiation by looking at a few examples.

Example 1. .3


  1. \begin{displaymath}
u = x^{2}\sin t,
\end{displaymath}


    \begin{displaymath}
\Rightarrow \qquad {\partial u \over \partial x} = 2x \sin t, \qquad
{\partial u \over \partial t} = x^{2} \cos t.
\end{displaymath}


  2. \begin{displaymath}
u = \left (x^{2} + t^{2}\right )^{2} + e^{x},
\end{displaymath}


    \begin{displaymath}
\Rightarrow \qquad {\partial u \over \partial x} = 2\left (...
...partial u \over \partial t} = 4t\left (x^{2} + t^{2}\right ).
\end{displaymath}


  3. \begin{displaymath}
u = \tan^{-1}\left ({x\over t}\right ),
\end{displaymath}


    \begin{displaymath}
\Rightarrow \qquad {\partial u \over \partial x} = {1\over 1 +
x^{2}/t^{2}}.{1\over t} = {t\over x^{2} + t^{2}},
\end{displaymath}


    \begin{displaymath}
\hbox{and }\qquad {\partial u \over \partial t} = {1\over 1 +
x^{2}/t^{2}}.{-x\over t^{2}} = {-x\over x^{2} + t^{2}}.
\end{displaymath}

If $u = u(r,s)$ and $r = r(x,t)$ and $s = s(x,t)$, then the chain rule gives

\begin{displaymath}
{\partial u\over \partial x} = {\partial u\over \partial r}...
...x} + {\partial u\over \partial s}{\partial s\over \partial x}.
\end{displaymath}


\begin{displaymath}
{\partial u\over \partial t} = {\partial u\over \partial r}...
...t} + {\partial u\over \partial s}{\partial s\over \partial t}.
\end{displaymath}

Example 1. .4 $u = \sin (r + s^{2})$ and $r = te^{x}$ and $s = x+t$. Hence,

\begin{displaymath}
{\partial r \over \partial x} = e^{x}t, \qquad {\partial s ...
...partial t} = e^{x},
\qquad {\partial s\over \partial t} = 1.
\end{displaymath}

and so

\begin{displaymath}
{\partial u\over \partial x} = \cos (r + s^{2}).te^{x} + \c...
... e^{x} + (x+t)^{2}\right ]\left \{t
e^{x} + 2(x+t)\right \},
\end{displaymath}


\begin{displaymath}
{\partial u\over \partial t} = \cos (r + s^{2}).e^{x} + \co...
...t e^{x} + (x+t)^{2}\right ]\left \{
e^{x} + 2(x+t)\right \}.
\end{displaymath}

This could have been obtained directly by substituting for $r$ and $s$ before differentiating so that $u = \sin \left ( t e^{x} + (x +
t)^{2}\right )$. Thus, for example, we get

\begin{displaymath}
{\partial u\over \partial t} = \cos \left [t e^{x} + (x+t)^{2}\right
]\left \{ e^{x} + 2(x+t)\right \},
\end{displaymath}

directly. However, the chain rule is more powerful and will be used in the theory of the second order wave equation.
next up previous
Next: Verification of Solutions to Up: Partial Differential Equations of Previous: Introduction
Prof. Alan Hood
2000-10-30