- #1

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(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?

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- Thread starter UrbanXrisis
- Start date

- #1

- 1,197

- 1

(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

- #2

Zurtex

Science Advisor

Homework Helper

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[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

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-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

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