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Let \begin{align}x&=at\\ y&=bt\\ z&=y^2\end{align}

Then $$\frac{\partial z}{\partial x} = ?$$

I solved this as

\begin{align}\frac{\partial z}{\partial x} &= 2y\frac{\partial y}{\partial x}\\\\ &= 2y\left(\frac{\partial y}{\partial t}\frac{\partial t}{\partial x}\right)\\\\ &= 2y\times b\times \frac 1a\end{align}

Is this correct?

In this case are $z$ and $x$ independent or dependent variables?

EDIT
If this is correct and the derivative is non zero, then $z$ and $x$ are dependent variable. Can you help me solve this bigger problem understanding partial derivatives in backpropagation algorithm

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    $\begingroup$ Just to check your answer note that $z=y^2=b^2t^2=\frac{b^2}{a^2}x^2$ so that $z_x=\frac{2b^2x}{a^2}$. $\endgroup$ – JP McCarthy May 24 '17 at 13:46
  • $\begingroup$ ... this agrees with your answer by the way. $\endgroup$ – JP McCarthy May 24 '17 at 14:02
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actually x, y, z all are in parametric form in t
but you can write z in term of x
therefore z and x are dependent variable because if they are independent
then the partial derivative of z with respect to x must be zero

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