1
$\begingroup$

I am learning chain rule for two variable functions and in Stewart's book i have the exercise:

If $z=f(x, y)$ and $x=r^2 + s^2, y = 2rs$ then get $\frac{\partial ^2z}{\partial r^2}$

And i can get $\frac{\partial z}{\partial r}$, but in development of the exercise in Stewart's book a term appears that I do not know where it comes from.

Where does the term $2\frac{\partial z}{\partial x}$ come from?

enter image description here

$\endgroup$
0
$\begingroup$

Answering question directly: it is first member from formula $(f\cdot g)'=f'g+g'f$. In your case $$\frac{\partial}{\partial r}\left(2r \cdot\frac{\partial z}{\partial x}\right)=\frac{\partial}{\partial r}2r \cdot \frac{\partial z}{\partial x}+ 2r \cdot\frac{\partial}{\partial r}\frac{\partial z}{\partial x} $$ and $\frac{\partial}{\partial r}2r = \frac{d}{d r}2r=2$

$\endgroup$
2
  • $\begingroup$ ohh, my bad. Thanks =) $\endgroup$ May 26 at 0:34
  • $\begingroup$ Glad to be helpful . $\endgroup$
    – zkutch
    May 26 at 0:36

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.