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\begin{document}

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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}




ANSWER

\begin{description}

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

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

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

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

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$\begin{array}{cc} \epsfig{file=160-2000-5.eps, width=35mm} \ & \
\epsfig{file=160-2000-6.eps, width=45mm}
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Domain $-\infty<y<\infty$\hspace{1.5cm}Range
$-\frac{\pi}{2}<x<\frac{\pi}{2}$

\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}&=&\frac{1}{1+(x^3)^2}.3x^2+\frac{(1+x^6)9x^2-3x^3(6x^5)}{(1+x^6)^2}\\
&=&\frac{3x^2}{1+x^6}+\frac{9x^2(1-x^6)}{(1+x^6)^2}\\
&=&\frac{3x^2\left\{1+x^6+3-3x^6\right\}}{(1+x^6)^2}\\
&=&\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*}
&&x^2\frac{d^2u}{dx^2}-4x\frac{du}{dx}+4u\\&=&
\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}x^3\\
&=&\frac{6x^4(2-x^6)-12x^4(1+x^6)}{(1+x^6)^2}\\
&=&\frac{6x^4(2-x^6-2-2x^6)}{(1+x^6)^2}\\
&=&\frac{6x^4(-3x^6)}{(1+x^6)^2}\\ &=&-\frac{18x^{10}}{(1+x^6)^2}
\end{eqnarray*}

\end{description}



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