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QUESTION

\begin{description}

\item[(a)]
If $f$ is a function of $u$ and $v$, where $u=x^2-y^2$ and $v=xy$,
use the chain rule to show that

$$\frac{\partial f}{\partial x}=2x\frac{\partial f}{\partial
u}+y\frac{\partial f}{\partial v},$$

and find the corresponding expression for
$\ds\frac{\partial^2f}{\partial y\partial x}$.

\item[(b)]
Use Maclaurin's theorem to show that

$$(1+x)^\frac{1}{3}=1+\frac{1}{3}x-\frac{1}{9}x^2+\frac{5}{81}x^3+R_3,$$

and write down the Lagrange form of $R_3$.

\end{description}

\bigskip

ANSWER

\begin{description}

\item[(a)]
$f=f(u,v)$

$\ds u=x^2-y^2,\ \frac{\partial u}{\partial x}=2x, \,
\frac{\partial u}{\partial y}=-2y,\hspace{1cm}v=xy,\
\frac{\partial v}{\partial x}=y, \, \frac{\partial v}{\partial
y}=x.$ $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial
u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial
v}\frac{\partial v}{\partial x}=2x\frac{\partial f}{\partial
u}+y\frac{\partial f}{\partial v}$$

\begin{eqnarray*}
&&\frac{\partial^2f}{\partial y
\partial x}=\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial
x}\right)\\ &&=2x\frac{\partial}{\partial y}\left(\frac{\partial
f}{\partial u}\right)+\frac{\partial f}{\partial
v}+y\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial
v}\right)\\ &&=2x\left\{\frac{\partial^2f}{\partial
u^2}\frac{\partial u}{\partial y}+\frac{\partial^2f}{\partial
v\partial u}\frac{\partial v}{\partial y}\right\}+\frac{\partial
f}{\partial v}+y\left\{\frac{\partial^2f}{\partial u\partial
v}\frac{\partial u}{\partial y}+\frac{\partial^2f}{\partial
v^2}\frac{\partial v}{\partial y}\right\}\\
&&=2x\left\{\frac{\partial^2f}{\partial
u^2}(-2y)+x\frac{\partial^2f}{\partial v\partial
u}\right\}+\frac{\partial f}{\partial
v}+y\left\{-2y\frac{\partial^2f}{\partial u\partial
v}+x\frac{\partial^2f}{\partial v^2}\right\}\\
&&\frac{\partial^2f}{\partial y\partial
x}=-4xy\frac{\partial^2f}{\partial
u^2}+2(x^2-y^2)\frac{\partial^2f}{\partial u\partial
v}+xy\frac{\partial^2f}{\partial v^2}
\end{eqnarray*}


\item[(b)]\

$\begin{array}{ll} f(x)=(1+x)^\frac{1}{3} &f(0)=1^\frac{1}{3}=1\\
f'(x)=\frac{1}{3}(1+x)^{-\frac{2}{3}}&
f'(0)=\frac{1}{3}(1)^{-\frac{2}{3}}=\frac{1}{3}\\
f''(x)=\frac{1}{3}(-\frac{2}{3})(1+x)^{-\frac{5}{3}}&
f''(0)=-\frac{2}{9}(1)^{-\frac{5}{3}}=-\frac{2}{9}\\
f'''(x)=-\frac{2}{9}(-\frac{5}{3})(1+x)^{-\frac{8}{3}}&
f'''(0)=\frac{10}{27}(1)^{-\frac{8}{3}}=\frac{10}{27}\\
f^{(4)}(x)=\frac{10}{27}(-\frac{8}{3})(1+x)^{-\frac{11}{3}}
\end{array}$

Substituting these into Maclaurins theorem gives

$\ds(1+x)^\frac{1}{3}=1+\frac{1}{3}x+\left(-\frac{2}{9}\right)\frac{x^2}{2!
}+\frac{10}{27}\frac{x^3}{3!}+R_3,\
\\
=1+\frac{1}{3}x-\frac{1}{9}x^2+\frac{5}{81}x^3+R_3,$

where $\ds
R_3=\frac{x^4}{4!}f^{(4)}(c)=\frac{x^4}{24}\left(-\frac{80}{81}\right)(1+c)^{-\frac{11}{3}}
=-\frac{10x^4}{243(1+c)^\frac{11}{3}}$

$0<c<x.$


\end{description}




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