# Tangent line

1. Oct 10, 2004

### 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?

2. Oct 10, 2004

### 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 $m = y'$

3. Oct 10, 2004

### 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?

4. Oct 10, 2004

### 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').