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I'm trying to calculate derivative for this function, and I'm stuck with two results. Can anyone help please?

My derivative:

$$z = \frac{y^3}{3} - \frac{2}{y^3} + \frac{y}{2}$$

The first result I get is $0$

The second is: $$y'(z) = \frac{-1}{\frac{-1}{2}- \frac{6}{y^2} - y^2}$$

I have no idea where the problem is, thanks.

Update, I have worked for about last 40 minutes on this solution:

$$z'(y) = \frac{1}{2}+ \frac{6}{y^4} + y^2$$

How does that look, I really hope better :D

user2570174
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  • A possible typo: the "x^2" should, I think, be "y^4". – Barry Cipra Dec 04 '13 at 20:05
  • Yea, sorry, it is y`(z) = - 1 / ( -( 1 / 2) - (6 / x^4 ) - y^2), is this the correct one? – user2570174 Dec 04 '13 at 20:07
  • I also think it should be a y instead of an x, not just a 4 instead of a 2. – Barry Cipra Dec 04 '13 at 20:10
  • Ok, I will try again. Sorry for inconvenience. – user2570174 Dec 04 '13 at 20:12
  • Also, can you explain what your reasoning is for the two results? I don't see how you got 0 for the first result. I'm also curious how you wound up with all the (unnecessary!) minus signs in the second. – Barry Cipra Dec 04 '13 at 20:14
  • http://rogercortesi.com/eqn/tempimagedir/eqn3797.png I got this, you were correct. I will start with stating that I'm very bad at math, and teacher gave me few samples, and I simply calculated the results based on those samples. – user2570174 Dec 04 '13 at 20:18

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The OP's updated solution looks quite nice (three cheers for learning TeX!). The original formula expresses $z$ as a function of $y$, so the derivative with respect to $y$, $z'(y)$, is correctly given (although it's slightly disconcerting to reverse the order of the powers of $y$). Assuming the initial question concerned implicit differentiation, it's more or less correct to write $$y'(z)={dy\over dz} = {1\over dz/dy} = {1\over z'(y)}$$ hence the OP's second result, modulo a typo, is also OK, although it's a mystery where all the minus signs came from. It's also a mystery (to me) how the OP got $0$ for a first result.

Barry Cipra
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