Expanding Brackets with Partial Derivatives

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SUMMARY

The discussion centers on the application of the product rule, specifically the Leibniz rule of differentiation, to expand the expression \( y\frac{\partial }{\partial z}(z\frac{\partial f}{\partial x}) \). The expanded form is confirmed as \( yz\frac{\partial^2 f}{\partial z \partial x} + y\frac{\partial f}{\partial x} \cdot 1 \), where the term \( \frac{\partial z}{\partial z} \) simplifies to 1. The user sought clarification on the steps to achieve this expansion, demonstrating a common challenge in calculus involving partial derivatives.

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  • Understanding of partial derivatives
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  • Knowledge of Leibniz notation for differentiation
  • Basic calculus concepts
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Homework Statement
Just need some help to fill in the gaps expanding partial derivative brackets
Relevant Equations
##(y\frac {\partial } {\partial z}(z\frac{\partial f} {\partial x}))##

##(yz\frac {\partial^2 f} {\partial z \partial x} + y\frac{\partial f} {\partial x} \frac{\partial z} {\partial z}))##
Hi,

I just need some (hopefully) quick calculus help.

I have the following:
##(y\frac {\partial } {\partial z}(z\frac{\partial f} {\partial x}))##

After it is expanded this is the solution:
##(yz\frac {\partial^2 f} {\partial z \partial x} + y\frac{\partial f} {\partial x} \frac{\partial z} {\partial z}))##

Could someone please let me know the steps to get to the solution?

Thanks
 
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It is the product or Leibniz rule of differentiation. The last term is ##\dfrac{\partial z}{\partial z}=1##.
 
Of course, thanks.

Its been a long day and I just couldn't see it. All I needed was a little nudge.

Thanks for the reply
 

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