Question

alright...now implicit differentiation of cos(x)sin(y)=x-y. show steps please! i'm trying to understand how to do this....my prof didn't explain it well

Answer 1

-sin(x)sin(y)+cos(y)cos(x)y'=1-y'
-sin(x)sin(y)-1=y'(-cos(y)cos(x)-1)
y'=tan(x)tan(y)+1

i believe its been a while since this but let me know

Answer 2

the calculus is correct

Answer 3

the algebra is not

Answer 4

2nd to 3rd line

Answer 5

\[y= \frac{\sin(x)\sin(y)+1}{\cos(x)\cos(y) +1} \neq \frac{ \sin(x)\sin(y)}{\cos(x)\cos(y)} + \frac{1}{1}\]

Answer 6

-sin(x)sin(y)-1=y'(-cos(y)cos(x)-1)
y'=+sin(x)sin(y)+1/(cos(y)cos(x)+1)

oops this should be the answer