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QUESTION

\begin{description}

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

$$\frac{\partial f}{\partial x}=\frac{1}{y}\frac{\partial
f}{\partial u}+y\frac{\partial f}{\partial v},$$ and find the
corresponding expression for $\ds\frac{\partial^2f}{\partial
x^2}$.

\item[(b)]
State the general form of Maclaurin's theorem with a remainder.
Use this to show

$$e^{2x}=1+2x+2x^2+R_2.$$

State the Lagrange form of the remainder $R_2$, and determine its
maximum value when $0\leq x\leq0.25$.

\end{description}


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ANSWER

\begin{description}

\item[(a)]
Using the chain rule $\ds\frac{\partial f}{\partial
x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial
x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}.$

$\ds\frac{\partial u}{\partial x}=\frac{1}{y},\ \frac{\partial
v}{\partial x}=y$, and hence

$\ds\frac{\partial f}{\partial x}=\frac{1}{y}\frac{\partial
f}{\partial u}+y\frac{\partial f}{\partial v}$

\begin{eqnarray*}
\frac{\partial^2f }{\partial x^2}&=&\frac{\partial }{\partial
x}\left(\frac{\partial f}{\partial x}\right)\\ &=&\frac{\partial
}{\partial x}\left(\frac{1}{y}\frac{\partial f}{\partial
u}\right)+\frac{\partial }{\partial x}\left(y\frac{\partial
f}{\partial v}\right)\\ &=&\frac{1}{y}\frac{\partial }{\partial
x}\left(\frac{\partial f}{\partial u}\right)+y\frac{\partial
}{\partial x}\left(\frac{\partial f}{\partial
v}\right)\\&=&\frac{1}{y}\bigg\{\frac{\partial^2f }{\partial
u^2}\underbrace{\frac{\partial u}{\partial
x}}_{=\frac{1}{y}}+\frac{\partial^2f }{\partial v\partial
u}\underbrace{\frac{\partial v}{\partial x}}_{=y}\bigg\}+y\bigg\{
\frac{\partial^2f }{\partial u\partial
v}\underbrace{\frac{\partial u}{\partial
x}}_{=\frac{1}{y}}+\frac{\partial^2f }{\partial
v^2}\underbrace{\frac{\partial v}{\partial x}}_{=y}\bigg\}\\
&=&\frac{1}{y^2}\frac{\partial^2f}{\partial
u^2}+\frac{\partial^2f}{\partial v\partial
u}+\frac{\partial^2f}{\partial u\partial
v}+y^2\frac{\partial^2f}{\partial v^2}\\
&=&\frac{1}{y^2}\frac{\partial^2f}{\partial
u^2}+2\frac{\partial^2f}{\partial u\partial
v}+y^2\frac{\partial^2f}{\partial v^2}
\end{eqnarray*}

\item[(b)]
Maclaurin's theorem:

$\ds
f(x)=f(0)+xf^{(1)}(0)+\frac{x^2}{2!}f^{(2)}(0)+\ldots+\frac{x^n}{n!}f^{(n)}(0)+R_n$

where $\ds R_n=\frac{x^{n+1}}{(n+1)!}f^{n+1}(\theta x),\
0<\theta<1$

\begin{tabular}{ll}
$f(x)=e^{2x}$&$f(0)=e^0=1$\\
$f^{(1)}(x)=2e^{2x}$&$f^{(1)}(0)=2e^0=2$\\
$f^{(2)}(x)=4e^{2x}$&$f^{(2)}(0)=4e^{0}=4$\\ $f^{(3)}(x)=8e^{2x}$
\end{tabular}

Substituting these values into Maclaurin's theorem,

$\ds e^{2x}=1+x(2)+\frac{x^2}{2!}(4)+R_2=1+2x+2x^2+R_2,\ R_2=\frac
{x^3}{3!}f^{(3)}(\theta x)\\ f^{(3)}(\theta x)=8e^{2\theta
x}<8e^\frac{1}{2},$ since $\ds0\leq2x\leq0.5$ and $0<\theta<1$\\
Hence $|R_2|<8e^\frac{1}{2}\left(\frac{x^3}{6}\right)
<\frac{4e^\frac{1}{2}}{3}\left(\frac{1}{4}\right)^3
=\frac{e^\frac{1}{2}}{48}=0.0343$

\end{description}



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