Partial Differentiation

Mar 2010
26
0
Malaysia
Find : ∂z/∂x
\(\displaystyle
z = 3x\sqrt{y} - xcos(xy)
\)

My solution:
\(\displaystyle Let K = xcos(xy)\)
\(\displaystyle Let u = x , v = cos(xy)\)
∂u/∂x = 1
∂v/∂x = \(\displaystyle -ysin(xy)\)
Product rule:
∂K/∂x = v(∂u/∂x) + u(∂v/∂x) = \(\displaystyle cos(xy) - xysin(xy)\)


∂z/∂x = \(\displaystyle 3\sqrt{y} -cos(xy) -xysin(xy)
\)
 
Last edited:
Jan 2010
564
242
Kuwait
There is no need to make substitutions ..
Fix your function please,,
 
Mar 2010
26
0
Malaysia
Sorry, im learning how to using Math equation...my final answer is ∂z/∂x =\(\displaystyle 3\sqrt{y} -cos(xy) -xysin(xy)\). Im just want to ask whether im correct or not.
 
Jan 2010
564
242
Kuwait
To write the square root of y : \sqrt{y}
You need the product rule to differentiate the second term,,
 
Mar 2010
26
0
Malaysia
To write the square root of y : \sqrt{y}
You need the product rule to differetiate the second term,,

Thanks. What you mean by "the product rule to differetiate the second term". (Worried)
 
Mar 2010
26
0
Malaysia
How can you differentiate the red one ?
Sorry...The question is ask to find ∂z/∂x for this question \(\displaystyle
z = 3x\sqrt{y} - xcos(xy)
\)

After differentiate xcos(xy), then i get\(\displaystyle
cos(xy) - xysin(xy)
\)
 
Jan 2010
564
242
Kuwait
I know,
You have:
\(\displaystyle z=3x\sqrt{y} - x \, cos(xy)\)

And you want to find \(\displaystyle z_x\) , Right ?
 
Mar 2010
26
0
Malaysia
I know,
You have:
\(\displaystyle z=3x\sqrt{y} - x \, cos(xy)\)

And you want to find \(\displaystyle z_x\) , Right ?
Is ∂z/∂x =\(\displaystyle z_x\) ?

If same, then yes.
 
Jan 2010
564
242
Kuwait
Yes.
They are the same.

\(\displaystyle z=3x\sqrt{y} - x \, cos(xy)\)

\(\displaystyle z_x=3\sqrt{y} - \left[ { \color{blue} cos(xy) - xy sin(xy) } \right] \)

I used the product in blue,,
 
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