1. ## Partial Differentiation

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 = $\displaystyle 3\sqrt{y} -cos(xy) -xysin(xy)$

2. There is no need to make substitutions ..

3. 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.

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
$\displaystyle z = 3x\sqrt{y} - { \color{red} xcos(xy) }$
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 $\displaystyle z = 3x\sqrt{y} - xcos(xy)$

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

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

And you want to find $\displaystyle z_x$ , Right ?

9. Originally Posted by General
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.

10. 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,,

11. Originally Posted by General
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,,
Ouch, got the minus sign. The partial differentiation was not exactly same as differentiation topic, knew the concepts now.