If y=|cos x| then f'(3pi/4) is?

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If y=|cos x| then f'(3pi/4) is?

Mathematics
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\[\frac{-1}{\sqrt{2}}\]
it should be +ve 1/root2
no, i don't think

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Other answers:

the slope of the tangent is negative at the reqd point in the graph of f(x)=|cosx|
sorry, its 1/rt(2). u r right.
ok how did you get it
in the interval (pi/2,pi) the function is f(x)=+cosx so its derivative is sinx and at 3pi/4 its value is 1/rt(2). Observe that i had to find the sign of the function in the neighbourhood of the given point.
f(x)=-cos x in the intervl pi/2 to pi
yes... so the value is positive.
i mean when u plug pi/2 in f(x), f(x) comes positive
f(x)=-cosx in pi/2 to pi derivative of cosx is -sinx so f'(x)=sinx
ok..got it
bingo

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