Solving Graph Analysis: Find Derivatives & Turning Point

[Air]

1. Homework Statement :
A curve for for $z>-1$ is given by:

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

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


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

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

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


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 $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$.

 

[snipez90]

On the second line of your attempt, you have dz/dy = 2ze^(z^2) but the right hand side is the result of differentiating dy by dz. That should fix your third line.

 

[Air]

On the second line of your attempt, you have dz/dy = 2ze^(z^2) but the right hand side is the result of differentiating dy by dz. That should fix your third line.

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

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

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

 

[Air]

Is this correct:

$\frac{\mathrm{d}^2y}{\mathrm{d}z^2} = 2e^{z^2} + 4z^2e^{z^2}$

$\frac{\mathrm{d}^2x}{\mathrm{d}z^2} = -\frac{1}{(1+z)^2}$

$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4z^4e^{z^2} + 8z^3e^{z^2} + 6z^2e^{z^2} + 4ze^{z^2} + 2e^{z^2}$

 

[snipez90]

The dy/dx is correct. You have a slight typo on the second line (should be dy/dz), no biggie.

Anyways I've never encountered such a second derivative but I would assume you would then take d/dx [(1+z)(2ze^(z^2))] and implicitly differentiate. This could get ugly, so first take the constant 2 out and probably would be a good idea to distribute the ze^(z^2) factor and applying the product rule twice.

I think this way works, though now you do need dz/dx, which is just e^x. And then you could express everything in terms of z in the end. However I wouldn't doubt that there is a better way of doing this.

 

[Air]

Anyways I've never encountered such a second derivative but I would assume you would then take d/dx [(1+z)(2ze^(z^2))] and implicitly differentiate. This could get ugly, so first take the constant 2 out and probably would be a good idea to distribute the ze^(z^2) factor and applying the product rule twice.

I've done it by using $\frac{\mathrm{d}^2y}{\mathrm{d}z^2}$ and $\frac{\mathrm{d}^2x}{\mathrm{d}z^2}$ then finding $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$. Are my answer's posted above correct?

 

[snipez90]

I don't think that manipulation works for derivatives higher than the first, i.e. dy/dx = (dy/dz)/(dy/dz) but you have to differentiate the resulting expression with respect to x after that. Though I'm working out the final answer at the moment.

 

[snipez90]

OK, after taking d/dx [(1+z)(2ze^(z^2))] with the fact that dz/dx = e^x = e^[ln(z+1)] = z+1 and checking it, I got

$$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=$$

2[e^(z^2)](z+1)(2z^3 + 2z^2 + 2z +1)

 

[HallsofIvy]

$$\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)= \frac{d}{dx}(1+z)(2ze^{z^2})$$
$$= \frac{dz}{dx}(2ze^{z^2})+ z\frac{d(2ze^z)}{dx}$$