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Function: $f(x,y,z) = e^{-x} (x^2 + y^2 + z^2)$

I need to differentiate this equation for $f_x, f_y$ and $f_z$: $\nabla f = \langle f_x, f_y, f_z \rangle$

What I Got: \begin{align} f_x &= −e^{−x}(x^2+y^2+z^2)+e^{−x}(x^2+y^2+z^2)2x \\ f_y &= 2e^{-x}y \\ f_z &= 2e^{-x}z \\ \implies \nabla f &= \langle -e^{-x} (x-2), \: 2e^{-x}y, \: 2e^{-x}z\rangle \end{align}

Are these the correct differentiation?

Thank you in advance.

  • $f_y$ and $f_z$ are correct. for $f_x$ you must use the product rule since you have an $x$ term both inside and outside the bracket multiplying with each other – thesmallprint Mar 25 '19 at 23:50
  • @thesmallprint I did the product rule and edited the original post. Does it look right? –  Mar 26 '19 at 00:29
  • let $u=e^{-x}$, then what is $\frac{\partial u}{\partial x}$? similarly, let $v=x^2+y^2+z^2$, then what is $\frac{\partial v}{\partial x}$? then, use the product rule. since you are partially differentiating, the product rule will involve partial derivatives too. just so you have a solution to compare your answer to, here's what you should get $$f_x=2xe^{-x}-e^{-x}(x^2+y^2+z^2).$$ – thesmallprint Mar 26 '19 at 12:47

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No. You have to use product rule. $f_x=-e^{-x}(x^{2}+y^{2}+z^{2})+2xe^{-x}$, etc.

Kavi Rama Murthy
  • 311,013
  • @KavuRamaMurthy so I did the product rule and I got $-e^{-x} x^2 + e^{-x}2x$ would this be correct? –  Mar 26 '19 at 00:21
  • @user430574 No, when you differentiate $e^{-x}$ you have to keep $(x^{2}+y^{2}+z^{2})$ as it is. You cannot drop $y^{2}+z^{2}$, – Kavi Rama Murthy Mar 26 '19 at 00:32
  • okay, so I added the things that I dropped, this is what I got. $fx = -e^{-x}(x2+y2+z2) + e^{-x}(x2+y2+z2)2x$ –  Mar 26 '19 at 00:41
  • @user430574 In the second term you are differentiating $x^{2}+y^{2}+z^{2}$ partially w.r.t. $x$ and the derivative is $2x$, not $2x(x^{2}+y^{2}+z^{2})$ – Kavi Rama Murthy Mar 26 '19 at 05:25