Question

Find dy/dx by implicit differentiation (leave your answer in terms of x and y).
cosxsiny=1

Answer 1

do you know the product rule??

Answer 2

yes this is how far i got i think i got confused somewhere
i got
0=-sin(xy)+cos(xy)
i know this is not right

Answer 3

i know the correct answer is tan(x)tan(y) but im not sure what to do next

Answer 4

When you derive cos(x)*sin(y) with respect to x, you get

cos(x)*sin(y)

-sin(x)*sin(y) + cos(x)*cos(y)*y' ... Don't forget to use the chain rule

Answer 5

so after deriving both sides with respect to x, you go from

cos(x)*sin(y) = 1

to

-sin(x)*sin(y) + cos(x)*cos(y)*y' = 0

Answer 6

thanks @jim_thompson5910 i forgot how to do that...its been awhile...in calc 2 we dont even use it

Answer 7

np

Answer 8

your last step is to solve for y'

Answer 9

how do you get the y' at the end ?

Answer 10

the derivative of sin(y) is cos(y)*y' because you first take the derivative of the outer function sin(..) to get cos(..), then you derive the inner function y to get y' and you multiply it with the previous result

Answer 11

this is all with respect to x

Answer 12

oh and I'm sure you know, but y' is the same as dy/dx

Answer 13

thanks i did now that but thanks for your help

Answer 14

yw