anonymous
  • anonymous
Find the derivative of f(x) = cos x, a = pi/2
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@zepdrix i ddi it but can u check my answer?
zepdrix
  • zepdrix
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?
anonymous
  • anonymous
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

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zepdrix
  • zepdrix
Your teacher said there is an equation? :) That's not much detail lolol Your derivative looks correct at least.
anonymous
  • anonymous
yea she said try again there is an equation
anonymous
  • anonymous
let me show you how i wrote it first okay?
anonymous
  • anonymous
i mean the answer i wrote first
zepdrix
  • zepdrix
k
anonymous
  • anonymous
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.
zepdrix
  • zepdrix
Oh so like the previous problems, you're looking for an equation of the tangent line?
anonymous
  • anonymous
yes sorry i made a mistake
anonymous
  • anonymous
yes yes the equation of the tangent line
zepdrix
  • zepdrix
\[\large y=-x+b\] 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.
anonymous
  • anonymous
ohhh okay
anonymous
  • anonymous
so instead of 0 = 0(pi/2) + b what would i write
zepdrix
  • zepdrix
\[\huge (\color{royalblue}{\pi/2},\;\color{orangered}{0}) \qquad = \qquad (\color{royalblue}{x},\;\color{orangered}{y})\] 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....
anonymous
  • anonymous
okay
anonymous
  • anonymous
so 0 = (-1)(pi/2) + b ?
zepdrix
  • zepdrix
yes
anonymous
  • anonymous
okay
anonymous
  • anonymous
thx
zepdrix
  • zepdrix
np

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