Evaluate $dy/(d (theta)$ when $theta$ = $pi$ for $y= 2ln2 (theta) -cos^2 2(theta)$ ?

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Answer 1 · 2018-07-06T10:29:31

The answer is $=2/pi$

The function is

$y=f(theta)=2ln(2theta)-cos^2(theta)$

We need

$(lnx)'=1/x$

$(cosx)'=-sinx$

Therefore, the derivative is

$(dy)/(d theta)=f'(theta)=4*1/(2theta)-2costheta*(-sintheta)$

$=2/theta+2costhetasintheta$

And, when $theta=pi$

$f'(pi)=2/pi+2sin(pi)cos(pi)$

$=2/pi+0$

Evaluate dy/(d (theta)dy/(d (theta) when thetatheta = pipi for y= 2ln2 (theta) -cos^2 2(theta)y= 2ln2 (theta) -cos^2 2(theta) ?