Question

Implicit Differentiation:
x + cos (x+y) = 0

A) csc (x+y)-1
B) csc (x+y)
C) 1/(sin(x+y))
D) 1/(1-x^2)^.5
E) 1-sin(x)/sin(y)

I did the question twice, not getting any of the answers listed.

Answer 1

Show me what you did and I'll show you where you went wrong and help you out. =D

Answer 2

I have the right answer here, I know how to do it, but I can't help you if you don't work with me.

Answer 3

\[x+\cos(x+y)=0\]
So I took the derivative (implicitly), and my first step ended up with...
\[1-\sin(x+y)*1+y'\]
I think my confusion is stemming from the inside part of the sine function, I know I chain rule, but did I differentiate the inside correctly?

Answer 4

You did the differentiation correctly, but remember that it is (1+y') in parentheses and it's all equal to 0.

Now you just need to solve for y'.

Answer 5

So by simplification...
\[1-\sin(x+y)*(1+y')=0\]
\[\sin(x+y)*(1+y')=-1\]
\[(1+y')=\frac{-1}{sin(x+y)}\]
\[y'=\frac{-1}{sin(x+y)}-1\]
Or did I do something wrong on the way there?

Answer 6

You should have 1=sin(x+y)(1+y'), not a negative one.

Also, remember that 1/sin(x)=csc(x)