Your attachment is a bit small and hard to read, but here's what I
think it says. Please provide any necessary corrections.
9. Use implicit differentiation to compute [the] equation of the tangent line to the curve cos(y) + sin(y) - x at the point (-1, \(\displaystyle \pi\))
cos(y) * x + sin(y) * x = x
-sin(y) * y' + cos(y) * y' = 1
Let x = 1, y = \(\displaystyle \pi\)
\(\displaystyle -sin(\pi) * y'(-1) + cos(\pi) * y'(-1) = 1\)
0 * y'(-1) + 1 * y'(-1) = 1
-y'(-1) = 1
y'(-1) = -1
Equation of the tangent line:
\(\displaystyle \dfrac{y-\pi}{x-(-1)}=-1\)
\(\displaystyle y - \pi = -x - 1\)
\(\displaystyle y = -x - 1 + \pi\)
Assuming the above is correct, it seems to me like you're getting a bit confused about what the notation means. Your confusion seems to stem from this line:
\(\displaystyle -sin(\pi) * y'(-1) + cos(\pi) * y'(-1) = 1\)
Your statement that (0)(-1) + (-1)(-1) = 1 is correct, but that's not what the problem is asking for.
After you've solved for the value of y'(-1), that is the derivative of the function y(x) at the point x = -1, and found that y'(-1) = -1, you can then plug it back in to check. You've merely verified that the value solved for is correct. If we momentarily forget about the derivative aspect and sub in another variable, say u, we'll have:
\(\displaystyle -sin(\pi) * u + cos(\pi) * u = 1\)
If you solved that, you'd come to the conclusion that u = -1. And
that is your answer. At no point along the way did you ever say the original equation was equal to -1, only the variable u. Returning to the version with the derivative, hopefully you can see that it's the same principle at work. y'(-1) can be treated as a variable in this context because it's an expression that you don't know the value of. You have an equation containing that variable you can work with to find its value.
