The discussion revolves around finding the derivative of the equation 2 sin(x) cos(y) = 1, which involves implicit differentiation and the application of the product and chain rules in calculus.
The discussion is active with various interpretations of the differentiation process being explored. Some participants have provided guidance on the correct application of differentiation rules, while others are still questioning the validity of their approaches and results.
There is a repeated emphasis on ensuring the correct application of differentiation rules, and some participants express uncertainty about their calculations and the steps involved in reaching a solution.
use the product rule, f(x)g(y') + g(y)f(x')EnumaElish said:How do you find the derivative of f(x)g(y)?
you didn't do the product rule rightBuBbLeS01 said:use the product rule, f(x)g(y') + g(y)f(x')
It should be f(x)g'(y)y' + f '(x)g(y).BuBbLeS01 said:f(x)g(y') + f(x')g(y)
you still have 1 more stepBuBbLeS01 said:ok thank you!
[2 sinxcosx]'= 2 [(sin(x)' cos(x)+ sin(x) cos(x)']BuBbLeS01 said:Homework Statement
find the derivative of...
2 sinxcosy = 1
The Attempt at a Solution
(2 cosxcosy) * (cosy')
cos y' = -2 cosxcosy
y' = (-2 cosxcosy)/(cos)
I know that's not right but I am not sure where I am making the mistake.