Partial Differentiation

wkn0524

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

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General

There is no need to make substitutions ..

wkn0524

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.

General

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

wkn0524

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)

General

$$\displaystyle z = 3x\sqrt{y} - { \color{red} xcos(xy) }$$
How can you differentiate the red one ?

wkn0524

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$$\displaystyle cos(xy) - xysin(xy)$$

General

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

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

wkn0524

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.

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,,

wkn0524