Equation of Tangent Line to y=cosx at a=pi/4

[UrbanXrisis]

Write the equation of the tangent line to the cure y=cosx at a=pi/4

(y-y1)=m(x-x1)

cos(pi/4)=sqrt(2)/2
y'=-sinx=-sin(pi/4)=-sqrt(2)/2

(y-sqrt(2)/2)=m(x-pi/4)
y'=-sqrt(2)/2(x-pi/4)+sqrt(2)/2

My teacher circled the y' and took a point off. I know that y'=-sinx but what should be in its place?

 

[Zurtex]

y = mx + c is the tangent line.

y' = \frac{dy}{dx} which is the gradient of your line. So actually at x = a you will find that m = y'

 

[UrbanXrisis]

The answer for the tangent line is
-sqrt(2)/2(x-pi/4)+sqrt(2)/2

However, I wrote y'=-sqrt(2)/2(x-pi/4)+sqrt(2)/2

which my teacher marked off a point for the y'
I know that -sqrt(2)/2(x-pi/4)+sqrt(2)/2 is the tangent line, but what does it equal to?

 

[Sirus]

Because the tangent line is a function on its own, it could just be f(x) or y; you are looking for the equation of the tangent line, not just its slope (which is y').