Question

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Answer 1

sin^(-1)(x) = x+y, so y = sin^(-1)(x) - x. From there you just take the derivative. (I'll give you a hint: sin^(-1)(x) is the same as saying 1/sin(x)).

Answer 2

What? \(\sin^{-1}(x) \ne \dfrac{1}{\sin(x)} = \csc(x)\)

Answer 3

You mean to take the inverse, do not confuse it with the exponent to the power of negative 1

Answer 4

Crap, you're right. My bad. The rest of it is correct, though. Just ignore the parenthetical statement.

Answer 5

An alternative is to use implicit differentiation
\[\Large x = \sin(x+y)\]
\[\Large \frac{d}{dx}[x] = \frac{d}{dx}[\sin(x+y)]\]
\[\Large 1 = \cos(x+y)*\frac{d}{dx}[x+y]\]
\[\Large 1 = \cos(x+y)*(1+y \ ')\]
solve for `y'`

Answer 6

Nice one jim

Answer 7

i got 0

Answer 8

How?

Answer 9

wait no it 1-cos(x+y)/cos(x+y)