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?
tanx = sinx/cosx, but
(tanx)' = sec^2(x)
1 + .5(9/4) + 4 = 6.125
To find f'(pi/3), you need to compute the derivative of the function f(x) and then evaluate it at x = pi/3. Let's go through the steps to find the correct answer.
First, let's find the derivative of f(x) using the rules of differentiation:
f'(x) = (2cos(x)) + [(3sin^2(x))(cos(x))] + sec^2(x)
Now, we need to evaluate f'(x) at x = pi/3. Let's substitute pi/3 into the derivative expression:
f'(pi/3) = (2cos(pi/3)) + [(3sin^2(pi/3))(cos(pi/3))] + sec^2(pi/3)
To simplify this expression, we need to recall the values of sin(pi/3), cos(pi/3), and sec(pi/3):
sin(pi/3) = sqrt(3)/2
cos(pi/3) = 1/2
sec(pi/3) = 2
Now, substitute these values into the expression for f'(pi/3):
f'(pi/3) = (2(1/2)) + [(3(sqrt(3)/2)^2)(1/2)] + 2^2
Simplifying further:
f'(pi/3) = 1 + [(3(3/4))(1/2)] + 4
f'(pi/3) = 1 + (9/8) + 4
f'(pi/3) = 5/8 + 33/8
f'(pi/3) = 38/8
f'(pi/3) = 4.75
Therefore, the correct answer for f'(pi/3) is 4.75, not 6.125. It seems there was an error in your calculations after substituting the values.