Parameter representation of derivatives

Question:
$x$ and $y$ are given as follows:
\begin{eqnarray*}
x &=& e^t \cos t  \\
y &=& e^t \sin t
\end{eqnarray*}
Express $\frac{d^2y}{dx^2}$ in terms of $t$.

Answer:
Take the derivative with respect to $t$.
\begin{eqnarray*}
\frac{dx}{dt} &=& e^t (\cos t -\sin t)  \\
\frac{dy}{dt} &=& e^t (\cos t +\sin t)
\end{eqnarray*}
Therefore,
\[ \frac{dy}{dx} = \frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}
              = \frac{\cos t + \sin t}{\cos t - \sin t}\]
Now, $\frac{d^2y}{dx^2}$ is a little tricky.
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\frac{dy}{dx} = \frac{d}{dt}\frac{dy}{dx}\frac{dt}{dx} \]
Therefore,
\begin{eqnarray*}
\frac{d^2y}{dx^2} &=& \frac{d}{dt}\left(\frac{\cos t + \sin t}{\cos t - \sin t}\right)\frac{1}{e^t (\cos t -\sin t)}  \\
                                &=& \frac{(-\sin t + \cos t)(\cos t - \sin t)-(\cos t + \sin t)(-\sin t - \cos t)}{e^t(\cos t - \sin t)^3}  \\
                                &=& \frac{\cos^2 t - 2\cos t\sin t + \sin^2 t + \cos^2 t + 2\cos t\sin t + \sin^2 t}{e^t(\cos t - \sin t)^3}  \\
                                &=& \frac{2(\cos^2 t + \sin^2 t)}{e^t(\cos t - \sin t)^3}    \\
                                &=& \frac{2}{e^t(\cos t - \sin t)^3}  
\end{eqnarray*}

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