1. ## Partial Differentiation

Find : ∂z/∂x
$
z = 3x\sqrt{y} - xcos(xy)
$

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

∂z/∂x = $3\sqrt{y} -cos(xy) -xysin(xy)
$

2. There is no need to make substitutions ..

3. Sorry, im learning how to using Math equation...my final answer is ∂z/∂x = $3\sqrt{y} -cos(xy) -xysin(xy)$. Im just want to ask whether im correct or not.

4. To write the square root of y : \sqrt{y}
You need the product rule to differentiate the second term,,

5. Originally Posted by General
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".

6. Originally Posted by wkn0524
$
z = 3x\sqrt{y} - { \color{red} xcos(xy) }
$
How can you differentiate the red one ?

7. Originally Posted by General
How can you differentiate the red one ?
Sorry...The question is ask to find ∂z/∂x for this question $
z = 3x\sqrt{y} - xcos(xy)
$

After differentiate xcos(xy), then i get $
cos(xy) - xysin(xy)
$

8. I know,
You have:
$z=3x\sqrt{y} - x \, cos(xy)$

And you want to find $z_x$ , Right ?

9. Originally Posted by General
I know,
You have:
$z=3x\sqrt{y} - x \, cos(xy)$

And you want to find $z_x$ , Right ?
Is ∂z/∂x = $z_x$ ?

If same, then yes.

10. Yes.
They are the same.

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

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

I used the product in blue,,

11. Originally Posted by General
Yes.
They are the same.

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

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

I used the product in blue,,
Ouch, got the minus sign. The partial differentiation was not exactly same as differentiation topic, knew the concepts now.