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can some one solve it f(x) = arctanx/x....f`(0) = ? can some one solve it f(x) = arctanx/x....f`(0) = ? @Mathematics
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so arctanx/x is basically tan inverse/x
yeah
so do it that way
i have got 0...but im not sure if its correct or not
whats your value for x that you plugged in?
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f`(0) = ?
help me with this, how do i differentiate sin(\[\cos(\pi t/4)\]
yes that is the right answer for you
(sin(cos(pi*t/4))` = cos ( cos(pi*t/4)*(-sin(pi*t/4)
differentiate sin(cos(πt/4), let u=cos(πt/4) Dsinu=(cosu)D(cos(πt/4) =(cosu)(-sin(πt/4)(pi/4)Dt
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thats right
=(cos(cos(πt/4) ))(-sin(πt/4)(pi/4)Dt
i got the question wrong, its actually only \[\cos (pit/4) oops\]
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