Tangent line

  • #1
1,197
1
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?
 

Answers and Replies

  • #2
Zurtex
Science Advisor
Homework Helper
1,120
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[itex]y = mx + c[/itex] is the tangent line.

[tex]y' = \frac{dy}{dx}[/tex] which is the gradient of your line. So actually at [itex]x = a[/itex] you will find that [itex]m = y'[/itex]
 
  • #3
1,197
1
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
574
2
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').
 

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