Thread: Differentiation of trigonometry equations
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Old June 27th, 2009, 11:36 AM
Soroban Soroban is offline
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Hello, Greg!

You must be differentiating incorrectly . . .

Quote:
Given that: .y\:=\:\cos2x + \sin2x, .find \frac{d^2y}{dx^2}

and show that: .\frac{d^2y}{dx^2} + 4y \:=\: 0

We have: .\begin{array}{ccc}y &=& \cos2x + \sin2x \\ \\ [-3mm] \dfrac{dy}{dx} &=& \text{-}2\sin2x + 2\cos2x \\ \\[-3mm] \dfrac{d^2y}{dx^2} &=& \text{-}4\cos2x - 4\sin2x \end{array}


Therefore: .\begin{array}{ccccccc} \dfrac{d^2y}{dx^2} &=& \text{-}4\cos2x - 4\sin 2x \\ \\[-3mm] +4y\;\; &=& 4\cos2x + 4\sin2x \\ \hline \\[-4mm] \dfrac{d^2y}{dx^2} + 4y &=& 0\qquad\qquad\qquad \end{array}
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