The Question:

A curve for for $\displaystyle z>-1$ is given by:

$\displaystyle x=\ln(1+z), \ y=e^{z^2}$.

Find $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$ and $\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ in terms of $\displaystyle z$ and show that the curve has only one turning point and that this must be a minimum.

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The attempt at a solution:

$\displaystyle \frac{\mathrm{d}z}{\mathrm{d}x} = \frac{1}{1+z}$

$\displaystyle \frac{\mathrm{d}z}{\mathrm{d}y} = 2ze^{z^2}$

$\displaystyle \therefore \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{(1+z)(2ze^{z^2})}$

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My problem I have encountered:

Can you check that it is correct so far and provide tips on how to differentiate further to find $\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ and how I can find the turning point to prove that this must be a minimum.