Question

partial derivative of xysin^-1(yz) with respect to y.

Answer 1

ah. so you have to use chain rule too? not just product rule?

Answer 2

treat all other variables as constants.

using chain rule and skipping steps, simplifying early, ect, i get:

\[x(\sin^{-1}(yz) + \frac{ yz }{ \sqrt{1 - (yz)^{2}} })\]

Answer 3

sorry. i mean product rule, not chain rule.

Answer 4

Hmm. could do you use chain rule when there are 3 things to multply? x*y*sin^-1(yz)

Answer 5

meant product rule*

Answer 6

you could use product rule when you have multiple things, but in this case you only have two functions of y

Answer 7

how do you use the product rule is what i meant to say

Answer 8

the product rule for 3 + functions gets messy

Answer 9

for f(y)g(y) its f'(y)g(y) + f(y)g'(y)

Answer 10

Alright, i'll look up how to do it.

Answer 11

\[\frac{ d }{ dy } f(y)g(y) = f'(y)g(y) + f(y)g'(y)\]

where:
\[f'(y) = \frac{ d }{ dy } f(y)\]

Answer 12

our case we have:\[f(y) = xy\]
\[g(y) = \sin^{-1}(yz)\]

Answer 13

finding f'(y) and g'(y) before jumping in and plugging is helpful for harder functions

Answer 14

\[f'(y) = x\]
\[g'(y) = \frac{ z }{ \sqrt{1 - (yz)^{2}} }\]

Answer 15

i found a video that explained it nicely, you dont have to post anything more. thanks for clarifying

Answer 16

np. glad you know it :)

Answer 17

ty once again