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?

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

  • 3
    $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.