Question

If I needed to 'prove' using calculus that if:
f(x) = A sin^2(BX+C) + A cos^2(BX+C)
Then: f'(x) = 0
How would I go about that? Teacher said 'not' to use the rule: sin^2θ + cos^2θ = 1
Thanks a heap! This site is great for those tricky Math-C concepts!

Answer 1

If A, B, and C are constants, you just have to differentiate them term by term and make proper use of the chain rules. When I did it, it just canceled out (the cosine in the 2nd term flips the sign).

Answer 2

Here ya go!
\[f(x) = Asin ^{2}(Bx+C)+Acos ^{2}(Bx+C)\]
\[f'(x) = 2Asin(Bx+C)*\cos(Bx+C)*B+2Acos(Bx+c)*(-\sin(Bx+C))*B\]
\[= 2ABsin(Bx+C)\cos(Bx+C) - 2ABcos(Bx+C)\sin(Bx+C)\]
(by commutative property)
\[=0\]