Partial Derivatives:
u= xy(sin^-1(yz)) find the first partial derivative with respect to x
For the product rule how do I derive sin^-1(yz) in respect to x???

Question

Partial Derivatives: 
u= xy(sin^-1(yz)) find the first partial derivative with respect to x
For the product rule how do I derive sin^-1(yz) in respect to x???

kainui · Answer

y and z are independent of x correct? Then this is simply a constant.

jhannybean · Answer

Use the product rule here. $\ f'g + g'f$

anonymous · Answer

Yes. So, it would be 0. @Kainui

kainui · Answer

Well if you look at it, really y(sin^-1(yz)) is all the coefficient on x. So you don't even need to use a product rule here.

jhannybean · Answer

$$\large U_x = (xy)'(sin^{yz}) + (sin^{yz})'(xy)$$

jhannybean · Answer

Hmm I guess. I use the product rule as a safety net, just so I know when I take the derivative with respect to x I can clearly see what cancels out, and whatnot.

anonymous · Answer

So, the answer would be y(sin^-1(yz)) @Kainui

jhannybean · Answer

And darn it, I wrote the $\sin^{-1} (yz)$ a bit wrong <_<

kainui · Answer

Suppose you try to use the product rule, you'll get: $$\LARGE u=xy \sin^{-1}(yz)$$ $$\Large \frac{\partial u}{\partial x}=\frac{d}{dx}(x)*y \sin^{-1}(yz)+x \frac{d}{dx}(y \sin^{-1}(yz))$$ You get: $$\Large \frac{\partial u}{\partial x}=(1)*y \sin^{-1}(yz)+x* (0)$$ 

which is the correct answer still.

anonymous · Answer

@Kainui Oh ok. I get it!! Thank you!

kainui · Answer

Cool, glad I could help! =P