Problem source

\begin{questionparts}
\item If
\[{\mathrm f}(x)=\tan^{-1}x+\tan^{-1}\left(\frac{1-x}{1+x}\right),\]
find ${\mathrm f}'(x)$. Hence, or otherwise, find a simple expression for 
${\mathrm f}(x)$.
\item Suppose that $y$ is a function of $x$ 
with $0 < y < (\pi/2)^{1/2}$ and
\[x=y\sin y^{2}\]
for $0 < x < (\pi/2)^{1/2}$. Show that
(for this range of $x$)
\[\frac{{\mathrm d}y}{{\mathrm d}x}=
\frac{y}{x+2y^2\sqrt{y^{2}-x^{2}}}.\]
\end{questionparts}

Solution source

\begin{questionparts}
\item 
\begin{align*}
&& f(x)&=\tan^{-1}x+\tan^{-1}\left(\frac{1-x}{1+x}\right) \\
\Rightarrow && f'(x) &= \frac{1}{1+x^2} + \frac{1}{1+\l \frac{1-x}{1+x} \r^2} \cdot \l \frac{-2}{(1+x)^2}\r \\
&&&= \frac1{1+x^2}- \frac{2}{(1+x)^2+(1-x)^2} \\
&&&= \frac1{1+x^2} - \frac{2}{2+2x^2} \\
&&&= 0
\end{align*}

Therefore $f(x) = \begin{cases} c_1 & \text{if } x < -1 \\
c_2 & \text{if } x > -1 \end{cases}$

$f(0) = \tan^{-1} 0 + \tan^{-1} 1 = \frac{\pi}{4}$
$\lim_{x \to \infty} f(x) = -\frac{\pi}{2} + \tan^{-1} -1 = -\frac{3\pi}{4}$

therefore $f(x) = \begin{cases} -\frac{3\pi}{4}& \text{if } x < -1 \\
\frac{\pi}{4} & \text{if } x > -1 \end{cases}$

\item \begin{align*}
&& x &= y \sin y^2 \\
\Rightarrow && \frac{\d x}{\d y} &= \sin y^2 + 2y^2 \cos y^2 \\
\Rightarrow && \frac{\d y}{\d x} &= \frac{1}{\sin y^2+2y^2 \cos y^2} \\
&&&=\frac{1}{\frac{x}{y}+2y^2 \sqrt{1-\sin^2y^2}} \\
&&&= \frac{y}{x + 2y^3 \sqrt{1-\frac{x^2}{y^2}}} \\
&&&= \frac{y}{x+2y^2 \sqrt{y^2-x^2}}

\end{align*}
\end{questionparts}