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1. Homework Statement :
A curve for for [itex]z>-1[/itex] is given by:
[itex]x=\ln(1+z), \ y=e^{z^2}[/itex].
Find [itex]\frac{\mathrm{d}y}{\mathrm{d}x}[/itex] and [itex]\frac{\mathrm{d}^2y}{\mathrm{d}x^2}[/itex] in terms of [itex]z[/itex] and show that the curve has only one turning point and that this must be a minimum.
2. The attempt at a solution:
[itex]\frac{\mathrm{d}z}{\mathrm{d}x} = \frac{1}{1+z}[/itex]
[itex]\frac{\mathrm{d}z}{\mathrm{d}y} = 2ze^{z^2}[/itex]
[itex]\therefore \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{(1+z)(2ze^{z^2})}[/itex]
3. The problem I have encountered:
Can you check that it is corerct so far and provide tips on how to differentiate further to find [itex]\frac{\mathrm{d}^2y}{\mathrm{d}x^2}[/itex].
A curve for for [itex]z>-1[/itex] is given by:
[itex]x=\ln(1+z), \ y=e^{z^2}[/itex].
Find [itex]\frac{\mathrm{d}y}{\mathrm{d}x}[/itex] and [itex]\frac{\mathrm{d}^2y}{\mathrm{d}x^2}[/itex] in terms of [itex]z[/itex] and show that the curve has only one turning point and that this must be a minimum.
2. The attempt at a solution:
[itex]\frac{\mathrm{d}z}{\mathrm{d}x} = \frac{1}{1+z}[/itex]
[itex]\frac{\mathrm{d}z}{\mathrm{d}y} = 2ze^{z^2}[/itex]
[itex]\therefore \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{(1+z)(2ze^{z^2})}[/itex]
3. The problem I have encountered:
Can you check that it is corerct so far and provide tips on how to differentiate further to find [itex]\frac{\mathrm{d}^2y}{\mathrm{d}x^2}[/itex].