1. Maxima/ Minima

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.

2. Originally Posted by Air
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.

__________________
The attempt at a solution:
$\displaystyle \frac{\mathrm{d}{\color{red}x}}{\mathrm{d}{\color{ red}z}} = \frac{1}{1+z}$

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

$\displaystyle \therefore \frac{\mathrm{d}y}{\mathrm{d}x} = {\color{red}\frac{dy}{dz} \cdot \frac{dz}{dx}} = {\color{red}2 z e^{z^2} \, (1 + z)}$

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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.
Note the corrections (in red).

To find the turning point, solve $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = 0: ~ z = 0$ (since z > -1) ....

3. Originally Posted by Air and revised by Mr F (see post #2)
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.

__________________
The attempt at a solution:
$\displaystyle \frac{\mathrm{d}x}{\mathrm{d}z} = \frac{1}{1+z}$

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

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

__________________
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.
$\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d} x} \left[ \frac{\mathrm{d}y}{\mathrm{d}x} \right] = \frac{\mathrm{d}}{\mathrm{d} z} \left[ \frac{\mathrm{d}y}{\mathrm{d}x} \right] \cdot \frac{\mathrm{d}z}{\mathrm{d}x} = \, ....$

(where I used the chain rule).

Once you have $\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ you can use the second derivative test to test the nature of the stationary point at z = 0.

4. Originally Posted by mr fantastic
$\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d} x} \left[ \frac{\mathrm{d}y}{\mathrm{d}x} \right] = \frac{\mathrm{d}}{\mathrm{d} z} \left[ \frac{\mathrm{d}y}{\mathrm{d}x} \right] \cdot \frac{\mathrm{d}z}{\mathrm{d}x} = \, ....$

(where I used the chain rule).

Once you have $\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ you can use the second derivative test to test the nature of the stationary point at z = 0.
$\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2e^{z^2}(2z^3+2z^2+2z+1)(1+z)$ ?

5. Originally Posted by Air
$\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2e^{z^2}(2z^3+2z^2+2z+1)(1+z)$ ?
That's what I get too (and what are the chances we'd both be wrong ).

By the way ..... I hope you differentiated $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$ by re-writing it as $\displaystyle e^{z^2} (2z + 2z^2) \, ....$

6. Originally Posted by mr fantastic
That's what I get too (and what are the chances we'd both be wrong ).

By the way ..... I hope you differentiated $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$ by re-writing it as $\displaystyle e^{z^2} (2z + 2z^2) \, ....$
If you've got the same answer, it's bound to be correct.

Yes, differentiated very similarly but took out the constant $\displaystyle 2$.

So I had $\displaystyle u=e^{z^2}, \ v=z+z^2$ and then when derivative was found, just multiplied it by $\displaystyle 2$ in the end.