Question

Find the linearization of cos(2x) at pi/4. The answer was pi/2 - 2x but I got pi/4 -x?

Answer 1

what did you get when you evaluated cos(2x) at x=pi/4
and what did you get when you evaluated the derivative of cos(2x) at x=pi/4 ?

Answer 2

@amourette

Answer 3

when i evaluated it the first time, I got 0, and when I derived it I got -1

Answer 4

cos(2x) evaluated at x=pi/4 is 0 so you are right there
but
-2sin(2x) evaluated at x=pi/4 is not -1...it is -2
I think you just forgot to multiply by 2...

Answer 5

\[y=f'(\frac{\pi}{4})(x-\frac{\pi}{4})+f(\frac{\pi}{4}) \\ \text{ where } f(x)=\cos(2x) \\ \text{ and } f'(x)=-2 \sin(2x) \text{ not just } -\sin(2x) \text{ which I what I believe you put maybe }\]

Answer 6

why do you multiply it by 2?

Answer 7

oh maybe you don't know chain rule?

Answer 8

ohhhh that's probably why. so when you're deriving -sin(2x), the 2 goes on the outside?

Answer 9

\[f(x)=\cos(2x) \\ f'(x)=(2x)' \cdot (\cos)'(2x) \\ f'(x)=2 \cdot [-\sin(2x)]\]

Answer 10

derivative of inside * derivative of outside

Answer 11

I see my mistake thank you very much!

Answer 12

np