Question

Find the derivative of f(x) = cos x, a = pi/2

Answer 1

@zepdrix i ddi it but can u check my answer?

Answer 2

sure :O what are you suppose to do? Just find the derivative then plug \(\large a\) in?
Or does this have something to do with approximation?

Answer 3

okay my teacher said there is an equation, so here is what i did f(x) = cos (x) f'(x) = -sin(x) x = a = pi/2 so f'(pi/2) = -sin (pi/2) = -1

Answer 4

Your teacher said there is an equation? :) That's not much detail lolol

Your derivative looks correct at least.

Answer 5

yea she said try again there is an equation

Answer 6

let me show you how i wrote it first okay?

Answer 7

i mean the answer i wrote first

Answer 8

k

Answer 9

d/dx ( cos (x)) = -sin(x)
-sin (pi/2) = -1, so the slope is -1, y = mx + b --> y = =1x + b
cos (pi/2) = 0 so the coordinates pare are (pi/2, 0) so y = 0x + b, 0(pi/2) + b equals to 0, 0 = y = 0x + 0 = y = 0 for my final answer.

Answer 10

Oh so like the previous problems, you're looking for an equation of the tangent line?

Answer 11

yes sorry i made a mistake

Answer 12

yes yes the equation of the tangent line

Answer 13

\[\large y=-x+b\]
From here you plugged in your point \(\large \left(\pi/2,\;0\right)\).
\[\large 0=-\pi/2+b \qquad \qquad \rightarrow \qquad \qquad b=\pi/2\]
I think you plugged your \(\large 0\) into the wrong spot.

Answer 14

ohhh okay

Answer 15

so instead of 0 = 0(pi/2) + b what would i write

Answer 16

\[\huge (\color{royalblue}{\pi/2},\;\color{orangered}{0}) \qquad = \qquad (\color{royalblue}{x},\;\color{orangered}{y})\]
Plug them into here, match up the colors.\[\huge \color{orangered}{y}=(-1)\color{royalblue}{x}+b\]

I'm not sure why you're plugging a \(\large 0\) in for your -1....

Answer 17

okay

Answer 18

so 0 = (-1)(pi/2) + b ?

Answer 19

yes

Answer 20

okay

Answer 21

thx

Answer 22

np