# Derivative for energy

1. Feb 10, 2016

### Dousin12

1. The problem statement, all variables and given/known data
1/2mz^2 +mgh=mgh-zero , get g
3. The attempt at a solution
z= velocity

z^2=g(2h0-2h)

if i set z^2=a
2h0=b (nonvariable)
2h=c

a=g(b-c)
y'=-g
Can i then say that

dz^2/d2h = -g

I wonder if every step is correct, The writing inbetween is not very important! I mostly wonder about the math. That if i do the substitution it actually work that way! x0 is not a variable!

dv^2/d2h feels like second derivate, but i use it as a first derivate or something, that was makes me clueless

2. Feb 10, 2016

### haruspex

No, it's a first derivative. A second derivative would have a power of 2 on the d itself (in the numerator).
It is correct, though you can take the 2 outside the 'd', in the denominator: $\frac{d(v^2)}{2dh}$.
The d(v2) can also be simplified.

3. Feb 10, 2016

### Dousin12

2v/2dh?

But y-axis has to be z^2 for it to show -g as a graph and not 2z hmmm

I was sortof proving that if y-axis is z^2 and x-axis 2h it will represent

Checked with theory numbers and it shows 9,81 excaltly so it works

4. Feb 10, 2016

### haruspex

If the d(v2) form gives you all you need, that's fine. But it can also be written as 2vdv, giving $v\frac{dv}{dh}$.