<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand\ds{\displaystyle}
\usepackage{epsfig}
\begin{document}

\parindent=0pt

QUESTION

\begin{description}

\item[(i)]
Express $\tan x$ as a quotient of trigonometric functions and
hence, or otherwise, derive the derivative of $\tan x$ with
respect to $x$.

\item[(ii)]
State the domain and range of the function $x=\tan^{-1}y$ and
sketch its graph.

\item[(iii)]
Use the identity $1+\tan^2x=\sec^2x$ to show that

$$\frac{d}{dy}(\tan^{-1}y)=\frac{1}{1+y^2}.$$

\item[(iv)]
Obtain the first and second derivatives with respect to $x$ of the
function $u=x\tan^{-1}(x^3)$ and hence verify that $u$ satisfies
the equation

$$x^2\frac{d^2u}{dx^2}-4x\frac{du}{dx}+4u=-\frac{18x^{10}}{(1+x^6)^2}.$$

\end{description}

\bigskip


ANSWER

\begin{description}

\item[(i)]
$\ds\tan x=\frac{\sin x}{\cos x}$

Therefore\\ $\ds\frac{d}{dx}(\tan x)=\frac{\cos x\cos x-\sin
x(-\sin x)}{\cos^2 x}$

$\ds =\frac{\cos^2x+\sin^2x}{\cos^2x}=\frac{1}{\cos^2x}=\sec^2x$

\item[(ii)]
If $x=\tan^{-1}y$, then $y=\tan x$.

\begin{center}
\epsfig{file=167-2000-4.eps, width=35mm} \ \ \
\epsfig{file=167-2000-5.eps, width=45mm}
\end{center}

Domain $-\infty&lt;y&lt;\infty$\hspace{1.5cm}Range
$-\frac{\pi}{2}&lt;x&lt;\frac{\pi}{2}$

\item[(iii)]
$x=\tan^{-1}y,\ y=\tan x$

Therefore $\ds\frac{dy}{dx}=\sec^2x=1+\tan^2x=1+y^2$

Therefore $\ds\frac{dx}{dy}=\frac{1}{1+y^2},$ or
$\ds\frac{d}{dy}(\tan^{-1}y)=\frac{1}{1+y^2}$

\item[(iv)]
$\ds u=x\tan^{-1}(x^3)\\
\frac{du}{dx}=\tan^{-1}(x^3)+x\left\{\frac{1}{1+(x^3)^2}.3x^2\right\}
=\tan^{-1}(x^3)+\frac{3x^3}{1+x^6}$
\begin{eqnarray*}
\frac{d^2u}{dx^2}&amp;=&amp;\frac{1}{1+(x^3)^2}.3x^2+\frac{(1+x^6)9x^2-3x^3(6x^5)}{(1+x^6)^2}\\
&amp;=&amp;\frac{3x^2}{1+x^6}+\frac{9x^2(1-x^6)}{(1+x^6)^2}\\
&amp;=&amp;\frac{3x^2\left\{1+x^6+3-3x^6\right\}}{(1+x^6)^2}\\
&amp;=&amp;\frac{3x^2(4-2x^6)}{(1+x^6)^2}
\end{eqnarray*}

i.e.$\ds\frac{d^2u}{dx^2}=\frac{6x^2(2-x^6)}{(1+x^6)^2}$

\medskip

Hence,
\begin{eqnarray*}
&amp;&amp;x^2\frac{d^2u}{dx^2}-4x\frac{du}{dx}+4u\\&amp;=&amp;
\frac{6x^4(2-x^6)}{(1+x^6)^2}-4x\left\{\tan^{-1}x^3+\frac{3x^3}{1+x^6}\right\}
+4x\tan^{-1}\left(x^3\right)\\
&amp;=&amp;\frac{6x^4(2-x^6)-12x^4(1+x^6)}{(1+x^6)^2}\\
&amp;=&amp;\frac{6x^4(2-x^6-2-2x^6)}{(1+x^6)^2}\\
&amp;=&amp;\frac{6x^4(-3x^6)}{(1+x^6)^2}\\ &amp;=&amp;-\frac{18x^{10}}{(1+x^6)^2}
\end{eqnarray*}

\end{description}



\end{document}
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