What will be value of this differential. Kindly help and explain.

ssafdarpk said:
Hi Dan,

Here it is View attachment 32619

Best regards,
Safdar
Click to expand...
Are [imath]H_i^1[/imath] and [imath]H_i^2[/imath] not functions of x? If so then [imath]\dfrac{ \partial H_i }{ \partial x }[/imath] is correct.

[imath]\dfrac{ \partial }{ \partial H_i } \left ( \dfrac{ \partial H_i }{ \partial x } \right ) = ?[/imath]. Recall that [imath]\dfrac{ \partial ^2 }{ \partial x \partial y} = \dfrac{ \partial ^2}{ \partial y \partial x}[/imath].

-Dan
 
topsquark said:
Are [imath]H_i^1[/imath] and [imath]H_i^2[/imath] not functions of x? If so then [imath]\dfrac{ \partial H_i }{ \partial x }[/imath] is correct.

[imath]\dfrac{ \partial }{ \partial H_i } \left ( \dfrac{ \partial H_i }{ \partial x } \right ) = ?[/imath]. Recall that [imath]\dfrac{ \partial ^2 }{ \partial x \partial y} = \dfrac{ \partial ^2}{ \partial y \partial x}[/imath].

-Dan
Click to expand...
Thanks Dan!

Do you imply that

[math]\frac{\partial}{\partial H_i} \bigg(\frac{\partial H_i}{\partial x}\bigg)= \frac{\partial}{\partial x} \bigg(\frac{\partial H_i}{\partial H_i}\bigg)[/math]
if so then it will result in

[math]\frac{\partial}{\partial x} \bigg(\frac{\partial H_i}{\partial H_i}\bigg) = \frac{\partial}{\partial x} \big(1\big) = 0[/math]
 
ssafdarpk said:
Thanks Dan!

Do you imply that

[math]\frac{\partial}{\partial H_i} \bigg(\frac{\partial H_i}{\partial x}\bigg)= \frac{\partial}{\partial x} \bigg(\frac{\partial H_i}{\partial H_i}\bigg)[/math]
if so then it will result in

[math]\frac{\partial}{\partial x} \bigg(\frac{\partial H_i}{\partial H_i}\bigg) = \frac{\partial}{\partial x} \big(1\big) = 0[/math]
Click to expand...
Exactly.

-Dan
 
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