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OpenStudy (anonymous):

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!

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OpenStudy (anonymous):

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).

OpenStudy (anonymous):

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) \[=0\]

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