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