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

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onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0@zepdrix i ddi it but can u check my answer?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1sure :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?

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0okay 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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Your teacher said there is an equation? :) That's not much detail lolol Your derivative looks correct at least.

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0yea she said try again there is an equation

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0let me show you how i wrote it first okay?

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0i mean the answer i wrote first

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0d/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
 2 years ago
Best ResponseYou've already chosen the best response.1Oh so like the previous problems, you're looking for an equation of the tangent line?

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0yes sorry i made a mistake

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0yes yes the equation of the tangent line

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1\[\large y=x+b\] 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.

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0so instead of 0 = 0(pi/2) + b what would i write

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1\[\huge (\color{royalblue}{\pi/2},\;\color{orangered}{0}) \qquad = \qquad (\color{royalblue}{x},\;\color{orangered}{y})\] 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....

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0so 0 = (1)(pi/2) + b ?
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