\hfill \thepage} %} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \section{Ma 116 Lecture 3/1/00} \vspace{1pt} \subsection{Partial Derivatives} \vspace{1pt} Definition: The \emph{first partial derivative of }$f\left( x,y\right) $% \emph{\ with respect to }$x$\emph{\ }at a point $\left( x,y\right) ,$ denoted by $\frac{\partial f}{\partial x}$ or $f_{x},$ is defined to be \vspace{1pt} \begin{equation*} f_{x}=\dfrac{\partial f}{\partial x}=\lim_{\Delta x\rightarrow 0}\dfrac{% f\left( x+\Delta x,y_{0}\right) -f\left( x,y_{0}\right) }{\Delta x} \end{equation*} \vspace{1pt} provided this limit exists (is finite). \vspace{1pt} Similarly, the rate of change of $f$ along the line $x=x_{0}$ (that is, parallel to the $y-$axis), called the \emph{first partial derivative of }$% f\left( x,y\right) $ \emph{with respect to }$y,$ is defined to be \vspace{1pt} \begin{equation*} f_{y}=\dfrac{\partial f}{\partial y}=\lim_{\Delta y\rightarrow 0}\dfrac{% f\left( x,y+\Delta y\right) -f\left( x,y\right) }{\Delta y} \end{equation*} provided this limit exists. \vspace{1pt} \subsection{Calculation of Partial Derivatives} The actual calculation of a partial derivative is straight-forward. For example, to find $f_{x}$ the definition tells us to hold $y$ fixed and differentiate with respect to $x.$ Similarly, to get $f_{y}$ we hold $x$ fixed and differentiate with respect to $y.$ We now present some examples that illustrate how one finds first partial derivatives. \paragraph{Example:} Let $z=f\left( x,y\right) =100-2x^{2}+4y^{3}.$ Then $\dfrac{\partial f}{% \partial x}=f_{x}=-4x$ and $\dfrac{\partial f}{\partial y}=f_{y}=12y^{2}$ \vspace{1pt} \paragraph{Example:} Let $f(x,y)=3x^{4}y^{2}-2\sin \left( x+2y\right) +3\left( x+y^{3}\right) ^{4}.$ Then $f_{x}=12x^{3}y^{2}-2\cos \left( x+2y\right) +12\left( x+y\right) ^{3}$ and $f_{y}=6x^{4}y-4\cos \left( x+2y\right) +12\left( x+y\right) ^{3}\left( 3y^{2}\right) .$ \vspace{1pt} \paragraph{Example:} Let $z=f\left( x,y\right) =\cosh \left( \dfrac{y}{x}\right) .$ Find $\frac{% \partial f}{\partial x}$ and $\frac{\partial f}{\partial y}.$ (Note: $\cosh t=\dfrac{e^{t}+e^{-t}}{2}$ and $\sinh t=\dfrac{e^{t}-e^{-t}}{2}.$) Note that from the formulas for $\cosh t$ and $\sinh t,$ it is follows that the derivative of $\sinh t$ is $\cosh t$ and the derivative of $\cosh t$ is $% \sinh t.$ Thus \vspace{1pt} $f_{x}=\sinh \left( \dfrac{y}{x}\right) \dfrac{\partial }{\partial x}\left( \dfrac{y}{x}\right) =\left( -\dfrac{y}{x^{2}}\right) \sinh \left( \dfrac{y}{x% }\right) $ and $f_{y}=\left( \dfrac{1}{x}\right) \sinh \left( \dfrac{y}{x}% \right) $ \vspace{1pt} \paragraph{Example:} If $f\left( x,y\right) =\dfrac{1}{\sqrt{x^{2}+y^{2}}}e^{xy}\cos \left( x^{2}+y^{2}\right) ,$ use SNB to find $\dfrac{\partial f}{\partial x}$ and $% \dfrac{\partial f}{\partial y}.$ \vspace{1pt}$\dfrac{\partial }{\partial x}\left( \dfrac{1}{\sqrt{x^{2}+y^{2}}% }e^{xy}\cos \left( x^{2}+y^{2}\right) \right) =\allowbreak \qquad -e^{xy}% \dfrac{\left( \cos \left( x^{2}+y^{2}\right) \right) x-y\left( \cos \left( x^{2}+y^{2}\right) \right) x^{2}-y^{3}\cos \left( x^{2}+y^{2}\right) +2\left( \sin \left( x^{2}+y^{2}\right) \right) x^{3}+2\left( \sin \left( x^{2}+y^{2}\right) \right) xy^{2}}{\left( x^{2}+y^{2}\right) ^{\frac{3}{2}}}% \allowbreak $ $\dfrac{\partial }{\partial y}\left( \dfrac{1}{\sqrt{x^{2}+y^{2}}}e^{xy}\cos \left( x^{2}+y^{2}\right) \right) =\allowbreak \qquad e^{xy}\dfrac{-y\cos \left( x^{2}+y^{2}\right) +\left( \cos \left( x^{2}+y^{2}\right) \right) x^{3}+\left( \cos \left( x^{2}+y^{2}\right) \right) xy^{2}-2\left( \sin \left( x^{2}+y^{2}\right) \right) yx^{2}-2\left( \sin \left( x^{2}+y^{2}\right) \right) y^{3}}{\left( x^{2}+y^{2}\right) ^{\frac{3}{2}}}% \allowbreak $ \vspace{1pt} \subsection{Higher Order Partial Derivatives} \vspace{1pt}Given a function $f(x,y)$ then once we have $\frac{\partial f}{% \partial x}$ and $\frac{\partial f}{\partial y}$ we can their partial derivatives with respect to $x$ or $y.$ Thus \vspace{1pt} \begin{eqnarray*} \dfrac{\partial }{\partial x}\left( \dfrac{\partial f}{\partial x}\right) &=&% \dfrac{\partial ^{2}f}{\partial x^{2}}=\dfrac{\partial f_{x}}{\partial x}% =f_{xx} \\ \dfrac{\partial }{\partial y}\left( \dfrac{\partial f}{\partial x}\right) &=&% \dfrac{\partial ^{2}f}{\partial y\partial x}=\dfrac{\partial f_{x}}{\partial y}=f_{xy} \\ \dfrac{\partial }{\partial x}\left( \dfrac{\partial f}{\partial y}\right) &=&% \dfrac{\partial ^{2}f}{\partial x\partial y}=\dfrac{\partial f_{y}}{\partial x}=f_{yx} \\ \dfrac{\partial }{\partial y}\left( \dfrac{\partial f}{\partial y}\right) &=&% \dfrac{\partial ^{2}f}{\partial y^{2}}=\dfrac{\partial f_{y}}{\partial y}% =f_{yy} \end{eqnarray*} \vspace{1pt} Note: The two partials $f_{xy}$ and $f_{yx}$ are called the mixed second order partial derivatives of $f\left( x,y\right) .$ Keep in mind that $% f_{xy} $ is found by first differentiating $f$ with respect to $x$ and then respect to $y,$ whereas $f_{yx}$ is found by first differentiating $f$ with respect to $y$ and then with respect to $x.$ These two mixed-partials need not be equal. However, for most of the functions encountered in applications they are indeed equal so that it does not matter which differentiation we do first. \vspace{1pt} \paragraph{Example:} Find the second partials of $f\left( x,y\right) =x^{3}y^{4}-2x^{2}e^{y}.$ $f_{x}=3x^{2}y^{4}-4xe^{y}$ and $f_{y}=4x^{3}y^{3}-2x^{2}e^{y}.$ Thus $% f_{xx}=6xy^{4}-4e^{y}$ \ \ $f_{xy}=12x^{2}y^{3}-4xe^{y}$ \ $% f_{yx}=12x^{2}y^{3}-4xe^{y}$ \ and $f_{yy}=12x^{3}y^{2}-2x^{2}e^{y}.$ \vspace{1pt} \paragraph{Example:} Consider the function $f\left( x,y\right) =x^{y}.$ Find all of the second order partial derivatives of $f$. Recall that if $u=u\left( x\right) $ and $a$ is a constant, then $\frac{d}{dx% }a^{u}=a^{u}\ln a\frac{du}{dx},$ whereas $\frac{d}{dx}x^{r}=r\allowbreak x^{r-1}.$ Thus $f_{x}=yx^{y-1}$ and $f_{y}=x^{y}\ln x\left( 1\right) .$ Therefore, $% f_{xx}=y\left( y-1\right) x^{y-2},$ \ $f_{xy}=x^{y-1}+yx^{y-1}\ln x,$ \ $% f_{yx}=yx^{y-1}\ln x+x^{y}\left( \dfrac{1}{x}\right) =yx^{y-1}\ln x+x^{y-1}$ and $f_{yy}=x^{y}\left( \ln x\right) ^{2}.$ \vspace{1pt} \paragraph{Example:} Show that the function $f\left( x,y\right) =\sin \left( x-y\right) $ satisfies the equation $f_{xx}-f_{yy}=0.$ $f_{x}=\cos \left( x-y\right) $ and $f_{y}=-\cos \left( x-y\right) $ so $% f_{xx}=-\sin \left( x-y\right) $ and $f_{yy}=-\sin (x-y).$ This gives the result. \vspace{1pt} \subsection{Partial Derivatives of Functions of Three or More Variables} Thus far we have limited out discussion to partial derivatives of functions of two variables. We may easily extend our discussion to functions of three or more variables. \paragraph{Example:} Find the first partial derivatives of $% g(x,y,z)=10xy^{2}z^{4}+3x^{3}y-7xy^{4} $ $g_{x}=10y^{2}z^{4}+9x^{2}y-7y^{4}\qquad g_{y}=20xyz^{4}+3x^{3}-28xy^{3}\qquad g_{z}=40xy^{2}z^{3}$ \vspace{1pt} \paragraph{Example:} Let $h\left( x_{1},x_{2},x_{3},x_{4},x_{5}\right) =\sin \left( x_{1}+x_{2}+x_{3}+x_{4}-x_{5}\right) +e^{x_{1}x_{2}x_{3}x_{4}x_{5}}.$ Find $% h_{x_{5}}.$ \vspace{1pt} $h_{x_{5}}=-\cos \left( x_{1}+x_{2}+x_{3}+x_{4}-x_{5}\right) +\left( x_{1}x_{2}x_{3}x_{4}\right) e^{x_{1}x_{2}x_{3}x_{4}x_{5}}$ \subsection{Chain Rule for Partial Derivatives} Recall that if $y=f\left( u\right) $ and and $u=g\left( x\right) ,$ then \vspace{1pt} \begin{equation*} \dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=f^{\prime }\left( u\right) g^{\prime }\left( x\right) =f^{\prime }\left( g\left( x\right) \right) g^{\prime }\left( x\right) \end{equation*} \vspace{1pt} This is called the \emph{chain rule }for a function of one variable. \vspace{1pt} The chain rule may be extended to functions of two or more variables. Let $% z=f\left( x,y\right) $ and suppose that $x=g\left( r,s\right) $ and $% y=h\left( r,s\right) ,$ that is, $x$ and $y$ are functions of the variables $% r$ and $s.$ Then $z$ may be thought of as a function of $r$ and $s,$ since $% z=f\left( x,y\right) =f\left( g\left( r,s\right) ,h\left( r,s\right) \right) =F\left( r,s\right) .$ Then the chain rule for partial derivatives gives \vspace{1pt} \begin{eqnarray*} \dfrac{\partial z}{\partial r} &=&\dfrac{\partial z}{\partial x}\dfrac{% \partial x}{\partial r}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{% \partial r}=\dfrac{\partial F}{\partial r} \\ \dfrac{\partial z}{\partial s} &=&\dfrac{\partial z}{\partial x}\dfrac{% \partial x}{\partial s}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{% \partial s}=\dfrac{\partial F}{\partial s} \end{eqnarray*} \vspace{1pt} \paragraph{Example:} Let $z=f\left( x,y\right) =e^{xy}$ and $x=r\cos s$ and $y=r\sin s.$ We shall find $\frac{\partial z}{\partial r}$ and $\frac{\partial z}{\partial s}.$ Now $\frac{\partial x}{\partial r}=\cos s,$ $\frac{\partial x}{\partial s}% =-r\sin s,$ $\frac{\partial y}{\partial r}=\sin s,$ and $\frac{\partial y}{% \partial s}=r\cos s.$ Thus \vspace{1pt} \begin{eqnarray*} \frac{\partial z}{\partial r} &=&ye^{xy}\cos s+xe^{xy}\sin s=2r\sin s\cos se^{r^{2}\sin s\cos s} \\ \frac{\partial z}{\partial s} &=&ye^{xy}\left( -r\sin s\right) +xe^{xy}\left( r\cos s\right) =r^{2}e^{r^{2}\sin s\cos s}\left( \cos ^{2}s-\sin ^{2}s\right) \end{eqnarray*} \vspace{1pt} \end{document} %%%%%%%%%%%%%%%%%%%%% End /document/lec_3_1_00.tex %%%%%%%%%%%%%%%%%%%%