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In part of my project in mathematics, I had to differentiate this: $$y(t)=A_{1}cos(2t)+A_{2}cos(4t)+A_{3}cos(6t)$$

And as someone advised, its derivative should be something like this: $$y^{"}(t)=-4A_{1}cos(2t)-16A_{2}cos(4t)-36A_{3}cos(6t)$$

But I don't understand how come it is not this: $$y^{"}(t)=-2A_{1}sin(2t)-4A_{2}sin(4t)-6A_{3}sin(6t)$$

Because as I know the derivation of $cos(x)$ is $-sin(x)$ and so for example derivation of $cos(2t)$ would be $-sin(2t)\times D(2t) = -sin(2t)\times 2 = -2sin(2t)$

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That is because $y''$ stands for the second derivative. The first derivative is exactly as you stated.

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    Oh my...I didn't somehow noticed it, I thought it was first derivative. So sorry, my mistake. Thank you for clarification :) – 2017-02-09