tanx = sinx/cosx, but
(tanx)' = sec^2(x)
1 + .5(9/4) + 4 = 6.125
For f(x) = 2sinx + (sinx)^3 + tanx find f'(pi/3). Ok, so what I tried was...
f'(x) = 2cosx + cosx(3(sinx)^2) + (sinx/cosx)
pi/3 = (1/2, sqrt(3)/2)
therefore, 2(1/2) + 1/2(3(sqrt(3)/2)(sqrt(3)/2) + (sqrt(3)/2 (2/1))
1 + .5(3 (3/4)) + sqrt(3)
1 + .5(9/4) + sqrt(3)
= 4.982...
but the answer in my text is 6.125. Where did I go wrong?
1 answer